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Vraag 1 Verslag
The molecular shape and bond angle of water are respectively
Antwoorddetails
The shape of water molecule = Bent/ V- shaped
The bond angle of water = 104.5°/ 105°
Vraag 2 Verslag
An element Z contains 80% of \( {}^{16}_{8}\mathrm{Z} \) and 20% of \( {}^{18}_{8}\mathrm{Z} \). Its relative atomic mass is
Antwoorddetails
R.A.M of Z = 16(80100)+18(20100)
= 12.8+3.6
= 16.4
Vraag 3 Verslag
A solution X, on mixing with \( \mathrm{AgNO_3} \) solution gives a white precipitate soluble in aqueous \( \mathrm{NH_3} \), a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain
Vraag 4 Verslag
When the end alkyl groups of ethyl ethanoate are interchanged, the compound formed is
Antwoorddetails
The compound formed when the end alkyl groups of ethyl ethanoate are interchanged is ethyl propanoate. This is because ethyl ethanoate consists of two parts: the "ethyl" group and the "ethanoate" group. The ethyl group is a two-carbon chain, and the ethanoate group is a combination of a one-carbon chain and a carbonyl group (C=O) that is also attached to an oxygen atom. When the end alkyl groups are interchanged, the "ethyl" group is moved from the second carbon to the first carbon of the ethanoate group, and the "propanoate" group is formed. The "propanoate" group consists of a three-carbon chain and the carbonyl group. Therefore, the resulting compound is ethyl propanoate, which has a chemical formula of CH3CH2COOCH2CH3. This compound is commonly used as a flavoring agent and has a fruity odor reminiscent of pears.
Vraag 5 Verslag
The combustion of carbon(ii)oxide in oxygen can be represented by equation.
\(2\mathrm{CO} + \mathrm{O}_2\ ?\ 2\mathrm{CO}_2\)
Calculate the volume of the resulting mixture at the end of the reaction if \(50\mathrm{cm}^3\) of carbon(ii)oxide was exploded in \(100\mathrm{cm}^3\) of oxygen
Antwoorddetails
Vraag 7 Verslag
Which of the following properties increases from left to right along the period but decreases down the group in the Periodic Table?
I. Atomic Number ii. Ionization energy iii. Metallic character iv. Electron affinity
Antwoorddetails
Ionization energy and electron affinity increase across a period, and decrease down a group.
Vraag 8 Verslag
What volume of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.
[Na = 23, N = 14, O = 16]
Antwoorddetails
To calculate the volume of a solution, we need to use the formula: moles of solute = concentration x volume First, let's find the number of moles of sodium trioxonitrate (V) in 5g of the solute. The molar mass of NaNO3 is: Na = 23 N = 14 3 x O = 3 x 16 = 48 Molar mass = 23 + 14 + 48 = 85 g/mol The number of moles of NaNO3 in 5g is: moles = mass / molar mass = 5 / 85 = 0.0588 moles Now, we can use the formula above to find the volume of the solution: moles of solute = concentration x volume volume = moles of solute / concentration volume = 0.0588 moles / 0.100 M volume = 0.588 litres Therefore, the correct answer is 0.588 litres of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.
Vraag 9 Verslag
If the cost of electricity required to discharge 10g of an ion \(X^{3+}\) is N20.00, how much would it cost to discharge 6g of ion \(Y^{2+}\)?
[1 faraday = 96,500C, atomic masses are X = 27, Y = 24]
Antwoorddetails
X3+
+ 3e−
→
X
3F = 27g
xF = 10g
x3=1027⟹x=109F
109
F ≡
N20.00
1F is equivalent to x
1109=x20
910=x20⟹x=N18.00
1F is equivalent to N18.00.
Y2+
+ 2e−
→
Y
2F = 24g
xF = 6g
x = 6×224=12F
1F = N18.00
12
F = 12×N18.00
= N9.00
Vraag 10 Verslag
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar mass of the compound is 180. Find the molecular formula.
[H = 1, C = 12, O = 16]
Antwoorddetails
The molecular formula of a compound is determined by the number of atoms of each element present in the molecule. To find the molecular formula, we need to determine the number of atoms of each element in the compound. First, we convert the percent composition to grams. For example, 40.0% carbon means 40.0 g of carbon per 100 g of compound. Then we divide the number of grams of each element by the molar mass of each element. For example, 40.0 g of carbon divided by the molar mass of carbon (12 g/mol) gives us 3.33 mol of carbon. Next, we convert the number of moles of each element to the number of atoms by multiplying the number of moles by Avogadro's number (6.022 x 10^23 atoms/mol). Finally, we balance the numbers of atoms of each element by dividing them by the smallest number of atoms of all the elements and rounding to the nearest whole number. In this case, the smallest number of atoms is 2, which is the number of hydrogen atoms. So, we divide the number of atoms of carbon and oxygen by 2 to balance the numbers of atoms of all the elements. Therefore, the molecular formula of the compound is C6H12O6.
Vraag 11 Verslag
Which of the following is a physical change?
Antwoorddetails
A physical change refers to a change in a substance that does not result in a change in its chemical composition. Out of the options provided, freezing ice cream is a physical change. This is because when ice cream is frozen, it changes from a liquid state to a solid state without any chemical reaction occurring. Exposing white phosphorus to air is a chemical change, as it reacts with oxygen in the air to form a new substance, phosphorus oxide. Burning kerosene is also a chemical change, as it undergoes combustion to form new substances, such as carbon dioxide and water vapor. Dissolving calcium in water is a physical change, as it simply involves the physical mixing of two substances without any chemical reaction occurring. Therefore, the only option that is a physical change is freezing ice cream.
Vraag 13 Verslag
Which of the following alkaline metals react more quickly spontaneously with water?
Antwoorddetails
The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) are the second most reactive metals in the periodic table, and, like the Group 1 metals, have increasing reactivity in the higher periods. Beryllium (Be) is the only alkaline earth metal that does not react with water or steam, even if metal is heated to red heat. Additionally, beryllium has a resistant outer oxide layer that lowers its reactivity at lower temperatures.
Magnesium shows insignificant reaction with water, but burns vigorously with steam or water vapor to produce white magnesium oxide and hydrogen gas:
A metal reacting with cold water will produce metal hydroxide. However, if a metal reacts with steam, like magnesium, metal oxide is produced as a result of metal hydroxides splitting upon heating.
The hydroxides of calcium, strontium and barium are only slightly water-soluble but produce sufficient hydroxide ions to make the environment basic, giving a general equation of:
| Order of reactivity | Metal | Reactions with water or steam |
|---|---|---|
| most reactive | potassium (K) | very vigorous reaction with cold water |
| ↑ | sodium (Na) | vigorous reaction with cold water |
| ↓ | calcium (Ca) | less vigorous reaction with cold water |
| least reactive | magnesium (Mg) | slow reaction with cold water, vigorous with steam |
Vraag 14 Verslag
Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are
Antwoorddetails
The chemical constituent of soda lime used to prepare methane in the laboratory is Ca(OH)2 (calcium hydroxide) and NaOH (sodium hydroxide). Soda lime is a mixture of these two compounds. When sodium ethanoate (NaC2H3O2) is heated with soda lime, it undergoes a reaction known as the Kolbe's reaction, which produces methane gas (CH4) as one of the products. The reaction can be represented as follows: 2NaC2H3O2 + 2Ca(OH)2 → 2CH4 + 2NaOH + 2CaCO3 In this reaction, the sodium ethanoate reacts with the calcium hydroxide to form calcium acetate (Ca(C2H3O2)2) and sodium hydroxide. The calcium acetate then decomposes to produce methane gas and calcium carbonate (CaCO3), which is a solid precipitate. Therefore, the chemical constituents of soda lime used to prepare methane in the laboratory are calcium hydroxide (Ca(OH)2) and sodium hydroxide (NaOH).
Vraag 15 Verslag
At 27°C, 58.5g of sodium chloride is present in 250cm\(^3\) of a solution. The solubility of sodium chloride at this temperature is?
(molar mass of sodium chloride = 111.0gmol\(^{-1}\))
Antwoorddetails
Given the Mass of the salt = 58.5g
Volume = 250 cm3
= 0.25 dm3
Mass concentration = MassVolume
= 58.50.25
= 234 gdm−3
Solubility (in moldm−3
= 234111
= 2.11 moldm−3
≊
2.0 moldm−3
Vraag 16 Verslag
In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm\(^{3}\) of 1.25 molar tetraoxosulphate (vi) acid?
Antwoorddetails
Equation of reaction : 2NaOH + H2 SO4 → Na2 SO4 + 2H2 O
Concentration of a base, CB = 0.5M
Volume of acid, VA = 10cm3
Concentration of an acid, CA = 1.25M
Volume of base, VB = ?
Recall:
CAVACBVB=nAnB
... (1)
N.B: From the equation,
nAnB=12
From (1)
1.25×100.5×VB=12
12.50.5VB=12
25 = 0.5VB
VB = 50.0 cm3
Vraag 17 Verslag
When chlorine water is exposed to bright sunlight, the following products are formed
Vraag 18 Verslag
What technique is suitable for separating a binary solution of potassium chloride and potassium trioxochlorate (V)?
Antwoorddetails
Fractional crystallization is the most suitable technique for separating a binary solution of potassium chloride and potassium trioxochlorate (V). This is because fractional crystallization is a process that separates a mixture of substances based on their solubility in a solvent at a particular temperature. In this case, potassium chloride and potassium trioxochlorate (V) have different solubilities in a solvent such as water at different temperatures. By carefully controlling the temperature, the solubility of each compound can be selectively increased or decreased, allowing them to be separated by crystallization. The less soluble compound will form crystals first and can be separated from the more soluble compound, which remains in the solution. Therefore, fractional crystallization can be used to separate potassium chloride and potassium trioxochlorate (V) in a binary solution.
Vraag 19 Verslag
\(200\text{cm}^3\) of \(0.50\text{mol}/\text{dm}^3\) solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is
Antwoorddetails
To solve this problem, we need to use the concept of stoichiometry and the solubility product constant (Ksp) of calcium hydrogen trioxocarbonate (IV). First, we need to write the balanced equation for the reaction that occurs when the solution of calcium hydrogen trioxocarbonate (IV) is heated: Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g) From the balanced equation, we can see that 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate. Therefore, we need to determine the number of moles of calcium hydrogen trioxocarbonate (IV) in the solution: Number of moles = concentration x volume Number of moles = 0.50 mol/dm³ x 0.2 dm³ Number of moles = 0.1 mol Since 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate, the number of moles of calcium carbonate produced will also be 0.1 mol. Next, we need to use the solubility product constant (Ksp) of calcium carbonate to determine the maximum amount of solid that can be precipitated: Ksp = [Ca²⁺][CO3²⁻] Ksp = 3.3 x 10⁻⁹ (at 25°C) At the maximum amount of solid precipitated, all the calcium carbonate formed will have precipitated, and the concentration of calcium ions and carbonate ions will be equal. Therefore, we can assume that the concentration of calcium ions and carbonate ions is both x. Substituting into the Ksp expression: Ksp = x² 3.3 x 10⁻⁹ = x² x = 5.74 x 10⁻⁵ mol/dm³ The mass of calcium carbonate precipitated can now be calculated: Mass = number of moles x molar mass Mass = 0.1 mol x 100.1 g/mol Mass = 10.01 g Therefore, the maximum weight of solid precipitated is approximately 10 g. Note that this calculation assumes that all the calcium carbonate precipitated as a solid, which may not always be the case in a real-world experiment. Additionally, this calculation does not take into account any losses due to filtration or other experimental errors.
Vraag 20 Verslag
Which of the following could not be alkane?
Antwoorddetails
An alkane is a type of hydrocarbon with only single bonds between the carbon atoms. It follows the general formula CnH2n+2, where "n" is the number of carbon atoms in the molecule. To determine whether a molecule is an alkane or not, we can calculate its molecular formula and check if it fits the general formula of alkane. Out of the given options, the third one (C7H14) cannot be an alkane. To see why, let's use the general formula of alkane, which is CnH2n+2. For C7H14 to be an alkane, it should have 2n+2 = 2(7) + 2 = 16 hydrogen atoms. However, C7H14 has only 14 hydrogen atoms, which means it does not follow the general formula of alkane. Therefore, C7H14 cannot be an alkane. The other options are as follows: - C4H10: This is butane, which is an alkane with four carbon atoms. - C5H12: This is pentane, which is an alkane with five carbon atoms. - C8H18: This is octane, which is an alkane with eight carbon atoms. In summary, the molecule C7H14 cannot be an alkane because it does not follow the general formula of alkane, while the other options are all examples of alkanes.
Vraag 21 Verslag
\( \mathrm{SO_3} \) is not directly dissolved in water in the industrial preparation of \( \mathrm{H_2SO_4} \) by the contact process because
Antwoorddetails
Vraag 22 Verslag
A radioactive nucleus has a half-life of 20 years, starting with 100,000 particles, how many particles will be left exactly at the end of 40 years
Antwoorddetails
The half-life of a radioactive nucleus is the time it takes for half of its particles to decay. This means that after 20 years, 100,000 particles will become 50,000 particles. After 40 years, we can find the number of particles remaining by counting the number of half-lives that have passed. Since 40 years is double the half-life of 20 years, this means that two half-lives have passed, so the number of particles will be halved twice. Starting with 100,000 particles: - After 1 half-life (20 years), there will be 50,000 particles remaining. - After 2 half-lives (40 years), there will be 25,000 particles remaining. So, exactly at the end of 40 years, there will be 25,000 particles remaining.
Vraag 23 Verslag
If acidified Potassium Dichromate(VI) (\(\mathrm{K_2Cr_2O_7}\)) acts as oxidizing agent, color changes from
Antwoorddetails
Potassium Dichromate (VI), when it is acidified, acts as an oxidizing agent. When this happens, the color changes from orange to green. This is because the orange color of the potassium dichromate is due to the presence of Cr(VI) ions, which are oxidized to Cr(III) ions. The green color that is produced is due to the formation of chromium(III) ions. In this reaction, the dichromate ions are being oxidized, which means that they are losing electrons, and the chromium ions are being reduced, which means that they are gaining electrons. The transfer of electrons causes the color change from orange to green.
Vraag 24 Verslag
The heat of formation of ethene, \( \mathrm{C_2H_4} \) is 50 \( \mathrm{kJmol^{-1}} \), and that of ethane, \( \mathrm{C_2H_6} \) is -82\( \mathrm{kJmol^{-1}} \). Calculate the heat evolved in the process:
\[ \mathrm{C_2H_4 + H_2 \rightarrow C_2H_6} \]
Antwoorddetails
The heat evolved in a chemical reaction can be calculated by subtracting the heat of formation of the reactants from the heat of formation of the products. In this case, the reactants are ethene (C2H4) and hydrogen (H2), and the product is ethane (C2H6). The heat of formation of ethene is 50 kJ/mol and that of hydrogen is 0 kJ/mol (because hydrogen is a reference element). The heat of formation of ethane is -82 kJ/mol. So, the heat evolved in the reaction is given by: Heat evolved = (Heat of formation of products) - (Heat of formation of reactants) = (-82 kJ/mol) - (50 kJ/mol + 0 kJ/mol) = -82 kJ/mol - 50 kJ/mol = -132 kJ/mol. Therefore, the heat evolved in the process is -132 kJ.
Vraag 25 Verslag
The two ions responsible for hardness in water are
Antwoorddetails
The ions responsible for hardness in water are Ca2+ and/or Mg2+. Hardness in water refers to the presence of calcium and magnesium ions, which are commonly found in natural water sources such as rivers, lakes, and groundwater. These ions can react with soap to form insoluble compounds, reducing the effectiveness of soap and causing scaling in pipes and appliances. The hardness of water is often measured in terms of the concentration of calcium and magnesium ions, expressed as calcium carbonate equivalents (CaCO3).
Vraag 26 Verslag
Which of the following pairs cannot be represented with a chemical formula?
Antwoorddetails
The pair that cannot be represented with a chemical formula is air and bronze. Air is a mixture of several gases, primarily nitrogen (N₂) and oxygen (O₂), with small amounts of other gases such as argon (Ar), carbon dioxide (CO₂), and neon (Ne). Since air is a mixture and not a pure substance, it cannot be represented by a chemical formula. Bronze, on the other hand, is an alloy composed mainly of copper (Cu) and tin (Sn) with small amounts of other metals. The composition of bronze can vary depending on the specific alloy, but it can be represented by a chemical formula such as CuSn. Sodium chloride (NaCl) is a compound composed of sodium (Na) and chlorine (Cl) in a fixed ratio of 1:1, and it can be represented by a chemical formula. Similarly, copper (Cu) and sodium chloride (NaCl) can each be represented by a chemical formula. Cu is an element, so its chemical formula is simply its symbol, while NaCl is a compound with a fixed ratio of sodium and chlorine atoms. Caustic soda (sodium hydroxide, NaOH) and washing soda (sodium carbonate, Na₂CO₃) are both compounds that can be represented by chemical formulas. NaOH consists of one sodium atom, one oxygen atom, and one hydrogen atom, while Na₂CO₃ consists of two sodium atoms, one carbon atom, and three oxygen atoms.
Vraag 27 Verslag
The electronic configuration of element Z is \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^1\). What is the formula of the compound formed between Z and tetraoxosulphate (VI) ion.
Antwoorddetails
Z = 1s2
2s2
2p6
3s2
3p1
?
We have Z3+
and SO2?4
The reaction : Z3+
+ SO2?4
?
Z2
(SO4
)3
.
Vraag 28 Verslag
A synthetic rubber is obtained from the polymerization of
Antwoorddetails
A synthetic rubber is obtained from the polymerization of isoprene. Isoprene is a type of hydrocarbon that can be polymerized, or chemically joined together, to form long chains. This process is called polymerization, and the resulting material is called a polymer. When isoprene is polymerized, it forms a synthetic rubber, which is a type of polymer that is used in a wide range of products, including tires, hoses, and adhesives. Synthetic rubber offers several advantages over natural rubber, including improved durability and resistance to heat, ozone, and chemicals.
Vraag 29 Verslag
Which of the following is the best starting material for the preparation of oxygen? Heating of trioxonitrate (v) with
Antwoorddetails
Vraag 30 Verslag
Which of the following statements about catalyst is false?
Antwoorddetails
The false statement about catalysts is: "catalysts do not alter the mechanism of the reaction and never appear in the rate law." Catalysts are substances that speed up chemical reactions without being consumed in the process. They achieve this by reducing the activation energy needed for the reaction to occur. Enzymes are a type of biological catalysts. In a chemical reaction, a catalyst is not consumed and does not appear in the overall balanced equation. However, catalysts can alter the mechanism of a reaction by providing an alternative pathway with a lower activation energy. This alternative pathway can have a different rate-determining step, which means that the presence of the catalyst can change the rate law of the reaction. Therefore, the statement that catalysts do not alter the mechanism of the reaction and never appear in the rate law is false.
Vraag 31 Verslag
Hydrocarbons which will react with Tollen's reagent conform to the general formula
Vraag 32 Verslag
Which of the following pollutants will lead to the depletion of ozone layer?
Antwoorddetails
The pollutant that leads to the depletion of the ozone layer is chlorofluorocarbon (CFCs). CFCs are man-made chemicals that were widely used in the past as refrigerants, solvents, and propellants. When CFCs are released into the atmosphere, they rise into the stratosphere, where they come into contact with ozone molecules. The chlorine atoms in CFCs react with ozone, breaking apart the ozone molecules and causing a reduction in the overall amount of ozone in the stratosphere. This process continues until all of the ozone-depleting chlorine atoms have been depleted. The resulting decrease in ozone in the stratosphere leads to an increase in the amount of harmful ultraviolet radiation that reaches the Earth's surface, which can have negative impacts on human health and the environment.
Vraag 33 Verslag
Which two gases can be used for the demonstration of the fountain experiment?
Antwoorddetails
Two gases that can be used in the study of fountain experiment is ammonia gas and hydrogen chloride gas. The experiment introduces concepts like solubility and the gas laws at the entry level.
Vraag 34 Verslag
Consider the reaction: \( \mathrm{A + 2B(g) \rightleftharpoons 2C + D(g)} \) (\( \Delta H = +\mathrm{ve} \))
What will be the effect of decrease in temperature on the reaction?
Antwoorddetails
The effect of a decrease in temperature on the reaction will be that the rate of the backward reaction will increase. In a chemical reaction, the rate of the forward and backward reactions are determined by the activation energy required for each step and the temperature of the system. When the temperature is decreased, the rate of the reaction decreases, and the rate of the backward reaction increases. This shift in the rate of the backward reaction means that there will be a shift in the position of the equilibrium of the reaction. As the rate of the backward reaction increases, the concentration of the reactants will increase and the concentration of the products will decrease, leading to a decrease in the overall yield of the products. In this reaction, as ΔH (the change in enthalpy) is positive, which means that the reaction is endothermic. Endothermic reactions absorb heat from the surroundings to proceed, so a decrease in temperature will lead to a decrease in the rate of the forward reaction and an increase in the rate of the backward reaction. This shift in the rate of the backward reaction will shift the position of the equilibrium of the reaction to the left, leading to an increase in the concentration of the reactants and a decrease in the concentration of the products.
Vraag 35 Verslag
\( \mathrm{Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2} \)
The indicator most suitable for this reaction should have a pH equal to
Antwoorddetails
Methyl orange is the best indicator for the reaction with range 3.1 - 4.4.
Vraag 36 Verslag
Which of the following factors will speed up the rate of evolution of carbon (iv) oxide in the reaction below?
\(2HCl + CaCO_3 \rightarrow CaCl_2 + H_2O + CO_2\)
Antwoorddetails
The following factors increase a reaction rate
- Increase in concentration of reactants
- Increase in temperature
- Addition of catalyst
- Increase in the surface area of reactant(s)
Vraag 37 Verslag
In the reaction:
M + N → P
ΔH = +Q kJWhich of the following would increase the concentration of the product?
Antwoorddetails
Increasing the temperature would increase the concentration of the product, P. The reaction rate, or the speed at which the reaction occurs, is influenced by temperature. An increase in temperature raises the kinetic energy of the reacting molecules, making it easier for them to collide and react. This leads to a higher rate of reaction and a higher concentration of the product, P. Adding a suitable catalyst can also increase the reaction rate, but it does not directly affect the concentration of the product. Increasing the concentration of P does not affect the reaction itself, but is a result of the reaction having taken place. Decreasing the temperature would slow down the reaction rate and reduce the concentration of the product.
Vraag 39 Verslag
Which process(es) is/are involved in the turning of starch iodide paper blue-black by chlorine gas?
Antwoorddetails
The process involved in the turning of starch iodide paper blue-black by chlorine gas is option number 3: chlorine oxidizes the iodide ion to produce iodine which attacks the starch to give the blue-black color. When chlorine gas comes in contact with iodide ions on the starch iodide paper, it oxidizes the iodide ion to form iodine. The iodine that is produced in this reaction is then able to react with the starch present on the paper to form a blue-black complex. This blue-black complex is formed due to the arrangement of the starch molecules and the iodine atoms in a way that causes them to absorb light at a specific wavelength, giving the blue-black color. Therefore, the blue-black color that is observed on the starch iodide paper is due to the reaction between iodine and starch, which is made possible by the oxidation of iodide ions by chlorine gas.
Vraag 40 Verslag
A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm3 of CO2 and 54g of water. The hydrocarbon should be
Antwoorddetails
In the question above an Hydrocarbon combust to give CO2 and H20
Let Hydrocarbon be
CxHy + x+Y/4O2= xCO2 + Y/2H2O
Mass of C0=44g and H2O=18g
at STP vol= 22.4
Therefore, 1mole of CO2 contains 44g and 22.4dm³ at STP
1mole = 22.4dm³
xmole = 89.6dm³
Cross multiplying x=89.6/22.4 =4mole of CO2 produce
1mole of H2O = 18g
Xmole = 56g
Cross multiplying
X = 56/18 = 3mole of H20
Then....
CxHy + X + y/4O2 = 4CO2+ 3H2O
Balancing
C4H6 + 6O2 = 4CO2 + 3H2O
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