JAMB UTME - Chemistry - 1991
Question 1 Report
Which of the following ions is a pollutant in drinking water even in trace amounts?
Answer Details
The pollutant in drinking water, even in trace amounts, among the given ions is Hg2+. Mercury (Hg) and its compounds are highly toxic and can cause serious health problems, including neurological and developmental damage, even at low concentrations. Therefore, the presence of mercury in drinking water can be harmful to human health. The other ions listed, Ca2+, Mg2+, and Fe2+, are essential minerals that are generally safe to consume in drinking water, and they do not pose a significant health risk in trace amounts.
Question 2 Report
30cm3 of oxygen at 10 atmospheres pressure is placed in a 20 dm3 container. Calculate the new pressure if temperature is kept constant?
Question 3 Report
How many grams of bromine will be required to completely react with 10 g of propyne?(C = 12, H = 1, Br = 80)
Answer Details
C3H4 + 2Br2 - C3Br4 + 2H2
(36 + 4) (4 X 80)
40 g of C3H4 reacts with 320 g of Br
10 " " will react with
(10)/(40) x (320) /(1) = 80g
Question 4 Report
The elections or two atoms Y and Z are arranged in shes as shown above. The bond formed between the atoms of Y and Z is
Answer Details
Question 5 Report
Which of the following is arranged in order of increasing electronegativity?
Answer Details
Question 6 Report
Which of the following compounds can exist as geometric isomers?
Answer Details
Geometric isomers are compounds with the same molecular formula, the same connectivity of atoms, and different arrangements of their substituent atoms in space. For compounds to be able to exist as geometric isomers, they must have a carbon-carbon double bond or a carbon-carbon triple bond. In the options given, only 2-methylbut-2ene and but-2-ene have a carbon-carbon double bond, which means they can exist as geometric isomers. But-1-ene does not have a double bond, and CI-C-Br is not an organic compound and does not have a carbon-carbon bond, so they cannot exist as geometric isomers. Therefore, the answer is: - 2-methylbut-2ene - But-2-ene
Question 7 Report
Magnesium (IV) oxide is known to hasten the decomposition of hydrogen peroxide. Its main action is to?
Answer Details
Magnesium (IV) oxide works as a catalyst in the decomposition of hydrogen peroxide, which means it helps speed up the reaction without getting used up in the process. The main action of the catalyst is to lower the activation energy for the reaction, making it easier for the reactants to overcome the energy barrier and react with each other. This makes the reaction occur more quickly than it would without the catalyst. Therefore, the correct option is: "lower the activation energy for the reaction".
Question 8 Report
CH3-CH-CH3
CH2CI Is known as?
Answer Details
The first molecule is a three-carbon chain with two carboxylic acid functional groups attached to the first and second carbon atoms. The second molecule is a two-carbon chain with a chlorine atom attached to the first carbon atom. Using the IUPAC naming system, the name of the first molecule is "propanedioic acid" or "malonic acid" and the name of the second molecule is "1-chloro-2-methylpropane". Therefore, the answer is 1-chloro-2-methylpropane.
Question 9 Report
A quantity of air was passed through a weighed amount of alkaline pyrogallol. An increase in the weight of the pyrogallol would result from the absorption of?
Answer Details
Alkaline pyrogallol is a substance used to absorb oxygen from the air. When air is passed through it, the oxygen in the air is absorbed by the pyrogallol, causing an increase in its weight. Therefore, the correct answer to the question is oxygen. Nitrogen, neon, and argon are all gases that are present in the air, but they do not react with pyrogallol in this way.
Question 10 Report
The data in the table above shows the rate of reaction of nitrogen (ll) oxide with chlorine at 25oC It can be concluded that doubling the initial concentration of NO increase the rate of reaction by a factor of
Answer Details
3.0 x 10-5 x 4 = 1.2 x 10-4
.: The rate increase by a factor of 4
Question 11 Report
What volume of CO2 at s.t.p would be obtained by reacting trioxocarbonate (IV) with excess acid?(G.M.V at s.t.p = 22.4 dm3)
Answer Details
The balanced equation for the reaction between trioxocarbonate (IV) and an acid is: CO2(g) + 2H+(aq) → H2O(l) + CO2(g) From the equation, we can see that one mole of trioxocarbonate (IV) reacts with two moles of H+ to produce one mole of CO2. At standard temperature and pressure (STP), one mole of any gas occupies 22.4 dm3 (or 22,400 cm3). Since the question states that there is an excess of acid, we can assume that all the trioxocarbonate (IV) will react, meaning that we only need to consider the amount of CO2 produced from one mole of the compound. Therefore, the volume of CO2 at STP produced from one mole of trioxocarbonate (IV) is 22.4 dm3 (or 22,400 cm3). Therefore, the correct answer is option B: 22.40 cm3.
Question 12 Report
50 cm3 of sulphur (IV) oxide, 800 cm3 of ammonia, 450 cm3 of hydrogen chloride, 1.0 cm3 of at 15°C. Which of the following is suitable for demonstrating the fountain experiment?
Answer Details
The fountain experiment is a classic chemistry demonstration where a gas is produced in a container and bubbles up through a liquid, creating a "fountain" effect. To perform this experiment, you need a gas that is less dense than the liquid and can dissolve in it to form bubbles. Out of the given options, only ammonia and hydrogen chloride can dissolve in water to form bubbles that are less dense than the liquid, making them suitable for the fountain experiment. Carbon (IV) oxide and sulphur (IV) oxide do not dissolve in water to form bubbles, and sulphur (IV) oxide and hydrogen chloride may react with each other to produce a toxic gas, making them unsuitable for the fountain experiment. Therefore, option C, ammonia and hydrogen chloride, is the suitable choice for the fountain experiment.
Question 13 Report
Which of the following are mixtures?
i. Petroleum
ii. Rubber latex
iii. Vulcanizer's solution
iv. Carbon (II) sulphide
Answer Details
Question 14 Report
In two separate experiments, 0.36g and 0.71g of chlorine combined with a metal X to give Y and Z respectively. An analysis showed that Y and Z contain 0.20g and 0.40g of X respectively. The data above represents the law of ?
Answer Details
0.36g of Cl combines with 0.2g of x
1g of Cl will combine with 1×0.20.32 = 0.56
0.71g of U combines with 0.49g of x
1g of U will combine with 1×0.471 = 0.56
Ratio 0.56 : 0.56 = 1:1
Question 15 Report
NO(g) + CO(g) ↔ (1)/(2)N2(g) + CO2(g) ∆H = -89.3 kj.
What conditions would favour maximum conversion of nitrogen (II) oxide and carbon (II) oxide in the reaction above?
Answer Details
The given reaction is exothermic, which means it releases heat when the reaction proceeds. According to Le Chatelier's principle, increasing the pressure and decreasing the temperature favors the reaction that produces fewer moles of gas. In this case, the reaction produces 2 moles of gas (N2 and CO2) from 2 moles of gas (NO and CO). So, decreasing the temperature and increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, i.e., the forward reaction. Therefore, the conditions that would favor maximum conversion of nitrogen (II) oxide and carbon (II) oxide are low temperature and high pressure. Hence, the correct answer is: Low temperature and high pressure.
Question 16 Report
Which of the following can be obtained by fractional distillation?
Answer Details
Fractional distillation is a process used to separate a mixture of different substances with different boiling points. The substance with the lowest boiling point will vaporize first and can be collected as a separate fraction. By repeating the process, each component of the mixture can be obtained in a separate fraction. Out of the given options, nitrogen from liquid air can be obtained by fractional distillation because liquid air is a mixture of nitrogen, oxygen, and other gases, and they have different boiling points. By fractional distillation, nitrogen can be separated from the mixture. Sodium chloride from seawater, iodine from a solution of iodine in carbon tetrachloride, and sulfur from a solution of sulfur in carbon disulfide cannot be obtained by fractional distillation because they are not mixtures of substances with different boiling points. Sodium chloride is a compound, and iodine and sulfur are elements. To obtain them, other methods such as precipitation, filtration, or chemical reactions would be necessary.
Question 17 Report
Copper (II) tetraoxosulphate (IV) is widely used as a?
Answer Details
Copper (II) tetraoxosulphate (IV) is widely used as a fungicide. It is effective against a range of fungal infections in plants, including powdery mildew, downy mildew, and black spot. It works by disrupting the cell membranes of the fungi, leading to their death. In addition to its use in agriculture, copper (II) tetraoxosulphate (IV) is also used in the production of various chemicals and as a pigment in ceramics and glass.
Question 18 Report
Which of the following statements is TRUE of the complete hydrolysis of a glycerine by sodium hygroxide?
Answer Details
Question 20 Report
Which of the following equilibria is unaffected by a pressure change?
Question 21 Report
Which of the following, in clear solution, forms a white precipitate when carbon (IV) oxide is bubbled into it for a short time?
Answer Details
When carbon (IV) oxide is bubbled into a clear solution, it forms a white precipitate of calcium carbonate (CaCO3). Therefore, the correct answer is Ca(OH)2, which is calcium hydroxide.
Question 22 Report
Ethene when passed into concentrated H2SO4 is rapidly absorbed. The product is diluted with water and then warmed to produce?
Answer Details
When ethene (C2H4) is passed into concentrated sulfuric acid (H2SO4), the ethene molecule gets protonated by H+ ions from the acid. This leads to the formation of a carbocation intermediate. Next, water is added to the carbocation to form an alcohol. In this case, since the product is being warmed, the most likely product formed is ethanol (C2H5OH). The reaction can be represented by the following equation: C2H4 + H2SO4 → C2H5OH Therefore, the correct option is "ethanol."
Question 23 Report
Oxygen gas can be prepared by heating?
Answer Details
Yes, oxygen gas can be prepared by heating certain compounds. One of the most common ways to obtain oxygen gas is by heating potassium trioxonitrate (V) or what is commonly known as potassium nitrate (KNO3). When potassium nitrate is heated, it undergoes a chemical reaction that produces oxygen gas and potassium oxide (K2O). The balanced chemical equation for this reaction is: 2KNO3(s) → 2KNO2(s) + O2(g) Therefore, by heating potassium nitrate, oxygen gas can be produced. The other compounds listed, such as ammonium trioxonitrate (V), ammonium trioxonitrate (III), and manganese (IV) oxide, do not produce oxygen gas when heated.
Question 24 Report
Atoms of element X (with 2 electrons in the outer shell). combine with atoms of Y (with 7 electrons in the outer shell).
Which of the following statements is FALSE? The compound formed
Answer Details
The false statement is "contains X2+ ions". When atoms of X (with 2 electrons in the outer shell) combine with atoms of Y (with 7 electrons in the outer shell), they are likely to form an ionic bond. In this bond, the X atom loses 2 electrons to become a cation with a 2+ charge (X2+), and the Y atom gains one electron to become an anion with a 1- charge (Y-). Therefore, the compound formed contains Y- ions but not X2+ ions, making statement 3 false. The compound formed will have a formula of XY because the cation (X2+) and anion (Y-) combine in a 1:1 ratio to form an electrically neutral compound.
Question 25 Report
The electronic configuration of an elements is 1s22s22p63s23p3. How many unpaired electrons are there in the element?
Answer Details
Question 26 Report
The appropriate test paper to use in the above experiment is moist
Question 27 Report
Which of the following are the products of the reaction between CH3COOH and CI2 in sunlight?
Answer Details
Question 28 Report
Which of the following salts can be melted without decomposition?
Answer Details
The salt that can be melted without decomposition is sodium carbonate (Na2CO3). This is because sodium carbonate has a high melting point and does not decompose at its melting point. On the other hand, calcium carbonate (CaCO3), magnesium carbonate (MgCO3), and zinc carbonate (ZnCO3) decompose upon melting due to the release of carbon dioxide (CO2) gas. Therefore, these salts cannot be melted without decomposition.
Question 29 Report
A given quantity of a gas occupies a volume of 228 cm3 at a pressure of 750mm Hg. What will be its volume at atmospheric pressure?
Answer Details
We can use the combined gas law to solve this problem, which states that the product of pressure and volume is directly proportional to the product of the number of moles of gas and the absolute temperature. Since the number of moles of gas and the temperature are constant, we can use the following equation: P1V1 = P2V2 where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure (atmospheric pressure), and V2 is the final volume (what we want to find). Substituting the values given in the problem, we get: (750 mm Hg)(228 cm^3) = (760 mm Hg)(V2) Solving for V2, we get: V2 = (750 mm Hg)(228 cm^3) / (760 mm Hg) V2 = 225 cm^3 Therefore, the volume of the gas at atmospheric pressure would be 225 cm^3. The answer is option (B).
Question 30 Report
The ions X- and Y+ are isoelectronic, each containing a total of 10 electrons.
How many protons are in the nuclei of the neutral atoms of X and Y respectively?
Answer Details
X- means one excess electron, hence 9 protons while Y +/- means one deficit electron hence 11 protons X and Y = 9 and 11
Question 31 Report
MnO-4 + 8H+ + ne ? Mn++ + 4H2O. Which is the value of n in the reaction above?
Answer Details
MnO-4 + 8H+ + ne- ? Mn++ + 4H2O
Since Mn has 7 electrons on one side and 2 electrons on the other side. The value of ne- is 5.
Question 32 Report
A property of a colloidal dispersion which a solution does not have is?
Answer Details
A solution is a homogeneous mixture of two or more substances, while a colloidal dispersion is a heterogeneous mixture where particles are dispersed in a continuous medium. One property of a colloidal dispersion that a solution does not have is the Tyndall effect. The Tyndall effect is the scattering of light by colloidal particles, which makes the path of the light visible. In a solution, the particles are too small to scatter light, so the path of light is not visible. Homogeneity, osmotic pressure, and surface polarity are properties that both solutions and colloidal dispersions can have.
Question 33 Report
Which of the following metals can be prepared in samples by the thermal decomposition of their trioxonitrate (V) salts?
Answer Details
Question 34 Report
If a current of 1.5 A is passed for 4.00 hours through a molten tin salt and 13.3g of tin is deposited, what is the oxidation state of the metal in the salt?(Sn = 118.7, F = 96500 C mol-1)
Answer Details
Q = It = 1.5 x 4 x 60 x 60
13.3 g = 216000 C
118.7 g = (216000 x118.7) / (13.3) = 1927759.4 C
06500 C - 1 mole
1927759.4 C = (1927759.4) / (96500) = 1.998 = 2
Question 35 Report
Addition of aqueous ammonia to a solution of Zn++ gives a white precipitate which dissolves in an excess of ammonia because?
Answer Details
The addition of aqueous ammonia to a solution of Zn^++ gives a white precipitate of zinc hydroxide (Zn(OH)2), which dissolves in excess ammonia to form a colorless, water-soluble complex ion called tetraamminezinc(II) ([Zn(NH3)4]^(2+)). This happens because zinc hydroxide is an amphoteric substance, which means it can react as an acid or a base. In the presence of aqueous ammonia, it acts as an acid and forms the complex ion which is readily soluble in water. Therefore, the correct answer is "zinc forms a complex which is readily soluble in excess ammonia."
Question 36 Report
A sample of temporary hard water can be prepared in the laboratory by?
Question 37 Report
Which of the following equimolar solutions, Na2CO3 Na2 SO
4CI FeCI3 CH3COONa,
Have pH greater than 7?
Question 38 Report
Which of these is an acid salt?
Question 39 Report
Which of the following gases will rekindle a brightly glowing splint?
Answer Details
The gas that will rekindle a brightly glowing splint is oxygen, which is present in N2O (nitrous oxide). When a glowing splint is introduced into a gas that contains oxygen, the oxygen reacts with the glowing ember and reignites the splint. This test is commonly used to identify the presence of oxygen in a gas mixture. Therefore, the correct answer is N2O.
Question 40 Report
How many structural isomers can be written for the alkyl bromide, C4H9Br?
Question 41 Report
An iron ore is known to contain 70.0% Fe2O3. The obtain from 80kg of the ore is?(Fe = 56, O = 16)
Answer Details
The given iron ore contains 70.0% Fe2O3, which means that for every 100 grams of the ore, 70 grams is Fe2O3 and 30 grams is other substances. We are asked to determine the amount of Fe obtained from 80 kg of the ore. To solve this problem, we need to first calculate the amount of Fe2O3 in 80 kg of the ore. Since the ore contains 70.0% Fe2O3, the amount of Fe2O3 in 80 kg of the ore is: 70.0% × 80 kg = 0.70 × 80 kg = 56 kg Next, we need to calculate the amount of Fe in 56 kg of Fe2O3. The molecular weight of Fe2O3 is: 2(Fe) + 3(O) = 2(56 g/mol) + 3(16 g/mol) = 160 g/mol This means that 1 mole of Fe2O3 contains 2 moles of Fe. Therefore, the number of moles of Fe in 56 kg of Fe2O3 is: (56 kg) / (160 g/mol) × (2 mol Fe / 1 mol Fe2O3) = 0.7 mol Fe Finally, we can calculate the amount of Fe in 80 kg of the ore by multiplying the number of moles of Fe by the molar mass of Fe: 0.7 mol Fe × 56 g/mol = 39.2 kg Therefore, the amount of Fe obtained from 80 kg of the ore is 39.2 kg. Hence, the correct answer is option (B) 39.2 kg.
Question 42 Report
Which of the following represents the type of bonding present in ammonium chloride molecule?
Answer Details
Ammonium chloride (NH4Cl) is a compound composed of positively charged ammonium ions (NH4+) and negatively charged chloride ions (Cl-). The type of bonding present in a molecule is determined by the difference in electronegativity between the atoms in the molecule. Electronegativity is the ability of an atom to attract electrons towards itself. In ammonium chloride, nitrogen (N) has a higher electronegativity than hydrogen (H), and chlorine (Cl) has a higher electronegativity than nitrogen. This means that the bonding between hydrogen and nitrogen is covalent, where the electrons are shared between the two atoms. However, the bonding between ammonium and chloride is ionic, where the ammonium ion (NH4+) donates its electron to the chloride ion (Cl-) to form an ionic bond. Therefore, the type of bonding present in ammonium chloride is ionic and covalent. "Ionic and dative covalent," is the correct choice.
Question 43 Report
A certain liquid has a high boiling point. It is viscous, non-toxic, miscible with water being very hygroscopic. This liquid is most likely to be?
Answer Details
The liquid described as having a high boiling point, being viscous, non-toxic, miscible with water, and very hygroscopic is most likely to be which is CH2OHCHOHCH2OH. The key characteristics of the liquid can help us identify its chemical structure. A high boiling point indicates that the liquid has strong intermolecular forces, which are likely to be due to the presence of hydrogen bonds. Viscosity is also related to intermolecular forces; a highly viscous liquid indicates strong forces between its molecules. The liquid is also described as being miscible with water, meaning that it can mix with water in any proportion. This property suggests that the liquid contains polar groups that can form hydrogen bonds with water molecules. Additionally, the liquid is hygroscopic, which means that it has a strong affinity for water and can absorb moisture from the air. This property further supports the presence of polar groups in the liquid. By looking at the chemical structures of the options, we can see that only CH2OHCHOHCH2OH, contains multiple hydroxyl (-OH) groups, which are highly polar and capable of forming hydrogen bonds with water molecules. These hydroxyl groups also contribute to the liquid's high boiling point and viscosity, as they can form intermolecular hydrogen bonds with other molecules of the liquid. Therefore, based on the given properties and chemical structures of the options, we can conclude that the liquid in question is most likely to be CH2OHCHOHCH2OH.
Question 44 Report
One advantages of detergents over soap is that detergents?
Answer Details
Detergents have an advantage over soap because they are able to form soluble salts with hard water. Hard water contains dissolved calcium and magnesium ions which react with soap to form insoluble scum. Detergents, on the other hand, are able to form soluble salts with these ions and do not react to form scum. This makes detergents more effective for cleaning in hard water areas. The other options listed do not accurately describe an advantage of detergents over soap.
Question 45 Report
The solubility of copper (II) tetraoxosulphate (Iv) is 75g in 100g of water at 100°C and 25g in 100g of water at 30°C. What mass of the salt would crystallize, if 50g of copper (II) tetraoxosulphate (VI) solution saturated at 100°C were cooled to 30°C ?
Answer Details
The solubility of copper (II) tetraoxosulphate (IV) is given as 75g in 100g of water at 100°C and 25g in 100g of water at 30°C. This means that at 100°C, 100g of water can dissolve up to 75g of the salt while at 30°C, 100g of water can dissolve up to 25g of the salt. If 50g of copper (II) tetraoxosulphate (VI) solution saturated at 100°C were cooled to 30°C, some of the salt would crystallize out since the solubility of the salt in water decreases as the temperature decreases. First, we need to find out how much of the salt is already dissolved in the 50g solution. At 100°C, 100g of water can dissolve 75g of the salt. Therefore, in 50g of water, we can dissolve: (75g/100g) x 50g = 37.5g of copper (II) tetraoxosulphate (IV) This means that 37.5g of the salt is already dissolved in the 50g solution at 100°C. When the solution is cooled to 30°C, the solubility of the salt reduces to 25g/100g of water. This means that the maximum amount of salt that can be dissolved in 50g of water at 30°C is: (25g/100g) x 50g = 12.5g of copper (II) tetraoxosulphate (IV) Therefore, the amount of salt that would crystallize out is the difference between the amount of salt already dissolved in the solution and the maximum amount of salt that can be dissolved in the solution at 30°C: 37.5g - 12.5g = 25g So, 25g of copper (II) tetraoxosulphate (IV) would crystallize out when 50g of the saturated solution at 100°C is cooled to 30°C. Therefore, the answer is 14.3g.
Question 46 Report
50 cm3 of sulphur (lV) oxide, 800 cm3 of carbon (IV) oxide will respectively saturate 1.0cm3 of water at 15oC. Which of the following is suitable for demonstrating the fountain experiment?
Question 47 Report
1.1 g of CaCI2 dissolved in 50 cm3 of water caused a rise in temperature of3.4°C. The heat of reaction, ∆H for CaCI2 in kj per mole is?
(Ca = 40, CI = 35.5, specific heat of water is 4.18 jk-1
Answer Details
This is a problem in thermodynamics that involves the calculation of the heat of reaction (∆H) for the dissolution of CaCl2 in water. The equation for the dissolution reaction is: CaCl2 (s) + H2O (l) → Ca2+ (aq) + 2Cl- (aq) The given information is: mass of CaCl2 = 1.1 g, volume of water = 50 cm3, temperature rise = 3.4°C, specific heat of water = 4.18 J K-1 g-1, and atomic weights of Ca and Cl are 40 and 35.5, respectively. To calculate ∆H, we need to use the equation: ∆H = -q / n where q is the heat absorbed by the solution (water + CaCl2), and n is the number of moles of CaCl2 dissolved. First, we need to calculate the number of moles of CaCl2: n = mass / molar mass n = 1.1 g / (40 g/mol + 2 x 35.5 g/mol) = 0.010 mol Next, we need to calculate the heat absorbed by the solution: q = m x c x ΔT where m is the mass of the solution, c is the specific heat of the solution, and ΔT is the temperature rise. The mass of the solution can be calculated from the density of water: density = mass / volume mass = density x volume mass = 1 g/cm3 x 50 cm3 = 50 g Now we can calculate q: q = 50 g x 4.18 J K-1 g-1 x 3.4°C = 714 J Finally, we can calculate ∆H: ∆H = -714 J / 0.010 mol = -71.4 kJ/mol Therefore, the answer is -71.1 kJ, which is closest to.
Question 48 Report
A sample of a gas exerts a pressure of 8.2 atm when confined in a 2.93 dm3 container at 20°C. The number of moles of gas in the sample is?
(R = 0.082 litre atm / deg mole)
Answer Details
PV = nRT
2.93 x 8.2 = 0.082 x 293 xn
∴ n = (2.93 x8.2) / (0.082 x 293)
n = 1
Question 49 Report
Calculate the volume of carbon (IV)oxide measured at s.t.p, produced when 1 kg of potassium hydrogen trioxocarbonate (IV) is totally decomposed by heat?(G.M.V. at s.t.p 22.4 dm3, K = 39, O = 16, C = 12, H = 1)
Answer Details
The question is asking for the volume of carbon dioxide produced when 1 kg of potassium hydrogen trioxocarbonate (IV) is decomposed at standard temperature and pressure (STP). We can start by writing the balanced chemical equation for the decomposition of potassium hydrogen trioxocarbonate (IV): 2KHCO3(s) → K2CO3(s) + H2O(g) + CO2(g) From this equation, we see that for every 2 moles of potassium hydrogen trioxocarbonate (IV) decomposed, 1 mole of carbon dioxide is produced. The molar mass of KHCO3 is: 1(39) + 1(12) + 3(16) = 100 g/mol Therefore, 1 kg (1000 g) of KHCO3 is equal to: 1000 g / 100 g/mol = 10 mol Since 2 moles of KHCO3 produce 1 mole of CO2, the number of moles of CO2 produced is: 10 mol / 2 = 5 mol At STP, 1 mole of any ideal gas occupies 22.4 dm^3. Therefore, the volume of CO2 produced is: 5 mol x 22.4 dm^3/mol = 112 dm^3 So the answer is 112 dm^3. Therefore, the correct option is: - 112 dm^3
