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Frage 1 Bericht
F is an inorganic salt. Carry out the following exercises on F. Record your observations and identify any gas(es) evolved. State the conclusions you draw from the results of each test.
(a) Transfer F into a test tube and add dil. HCl to it until a clear solution is obtained.
(b) Divide the clear solution obtained from (a) into two portions
(i) To the first portion add NaOH\(_{(aq)}\) in drops and then in excess.
(ii) To the second portion, add few drops of potassium hexacyanoferrate (II) solution.
F = (NH\(_4\))\(_2\)CO\(_3\)
| TEST | OBSERVATION | INFERENCE | |
| (a) | F + t.t + dil. HCl | Effervescence of colourless, odourless gas. Evolved gas turned lime water milky |
CO\(_2\) from CO\(_3\)\(^{2-}\) or HCO\(_3\) confirmed. |
| b(i) | 1st portion + t.t + NaOH in drops and
then in excess + heat to warm |
No visible reaction, No ppt was also formed Effervescence of a colourless gas with an irritating smell. Gas turned red litmus to blue,also produced white fumes with conc. HCl |
NH\(_3\) from NH\(_4\)\(^+\) confirmed. |
| (ii) | 2nd portion + t.t + few drops of potassium hexacyano ferrate(II) solution | No ppt formed and no visible reaction was observed | Fe\(^{2+}\)/ Fe\(^{3+}\) absent |
Antwortdetails
F = (NH\(_4\))\(_2\)CO\(_3\)
| TEST | OBSERVATION | INFERENCE | |
| (a) | F + t.t + dil. HCl | Effervescence of colourless, odourless gas. Evolved gas turned lime water milky |
CO\(_2\) from CO\(_3\)\(^{2-}\) or HCO\(_3\) confirmed. |
| b(i) | 1st portion + t.t + NaOH in drops and
then in excess + heat to warm |
No visible reaction, No ppt was also formed Effervescence of a colourless gas with an irritating smell. Gas turned red litmus to blue,also produced white fumes with conc. HCl |
NH\(_3\) from NH\(_4\)\(^+\) confirmed. |
| (ii) | 2nd portion + t.t + few drops of potassium hexacyano ferrate(II) solution | No ppt formed and no visible reaction was observed | Fe\(^{2+}\)/ Fe\(^{3+}\) absent |
Frage 2 Bericht
(a) An aqueous solution of a salt was added to excess sodium sodium trioxocarbonate (IV) solution in a test tube. There was effervescence, the test tube became warm and a white precipitate was observed. State three inferences that could be drawn from these observations.
(b) Consider the following salts; NH\(_4\)Cl; PbSO\(_4\); NaHCO\(_3\) and Cu(NO\(_3\))\(_2\).
Select from the list, the salt(s) which:
(i) would not readily dissolve in water;
(ii)produce(s) effervescence with dilute mineral acids;
(iii) decompose(s) on heating;
(iv) dissolve(s) in water to form an alkaline solution;
(v) sublime(s) on heating
(c) A solution of an acid was titrated with sodium hydroxide solution using methylorange as the indicator. State three precautions that are necessary when carrying out such an experiment.
a(i) Salt solution is acidic
- Reaction is exothermic.
- White precipitate is likely to be a trioxocarbonate(IV) salt of Al, Ca and Zn.
- Prepared solution contained soluble salt of Al, Ca and Zn.
b(i)PbSO\(_4\)
(ii) NaHCO\(_3\)
(iii) NaHCO\(_3\) ; Cu(NO\(_3\))\(_2\).
(iv) NaHCO\(_3\)
(v) NH\(_4\)Cl
(c) Precautions to be taken during titration include:
- Wear a lab coat to protect your clothing from spills.
- Ensure to rinse the burette with an acid solution
- Also ensure the funnel is removed immediately after filling the acid into the burette.
- Avoid error due to parallax while taking the reading to determine the end point.
- Ensure all glassware is clean and free of bubbles.
Antwortdetails
a(i) Salt solution is acidic
- Reaction is exothermic.
- White precipitate is likely to be a trioxocarbonate(IV) salt of Al, Ca and Zn.
- Prepared solution contained soluble salt of Al, Ca and Zn.
b(i)PbSO\(_4\)
(ii) NaHCO\(_3\)
(iii) NaHCO\(_3\) ; Cu(NO\(_3\))\(_2\).
(iv) NaHCO\(_3\)
(v) NH\(_4\)Cl
(c) Precautions to be taken during titration include:
- Wear a lab coat to protect your clothing from spills.
- Ensure to rinse the burette with an acid solution
- Also ensure the funnel is removed immediately after filling the acid into the burette.
- Avoid error due to parallax while taking the reading to determine the end point.
- Ensure all glassware is clean and free of bubbles.
Frage 3 Bericht
C is a mixture of two simple inorganic salts, one of which is a sodium salt.
Perform the following exercises on C. Record your observation and identify any gas(es) evolved. State the conclusions you draw from the results of each test.
(a) Put all of C into a test tube and add about 10 cm\(^3\) of distilled water and shake.
b(i) Put about 2 cm\(^3\) of the filtrate into a boiling tube and heat strongly.
(ii) Put half of the residue into a test tube and add dil. HCl.
(iii) To about 2 cm\(^3\) of the clear solution from b(ii), add aqueous NaOH in drops, then in excess.
(iv) To another 2 cm\(^3\) of the clear solution from b(ii), add aqueous ammonia in drops, then in excess.
| Test | Observation | Inference | |
| (a) | C + 10cm\(^3\) of H\(_2\)O | C partially dissolves | C is a mixture of a soluble and insoluble salt. |
| b(i) | 2cm\(^3\) of filtrate + heat | Colourless and odourless gas in form of bubbles | CO\(_3\)\(^{2-}\) suspected |
| (ii) | residue + t.t + dil. HCl | No visible reaction | NO\(_3\)\(^-\), SO\(_4\)\(^{2-}\) suspected |
| (iii) | Solution from b(ii) + NaOH in drops and then in excess | White ppt which is dissolves in excess NaOH | Zn\(^{2+}\) , Pb\(^{2+}\) , Al\(^{3+}\) suspected |
| (iv) | Solution from b(ii) + NH\(_3\) in drops and then in excess | White ppt insoluble in excess NH\(_3\) | Pb\(^{2+}\) confirmed. |
Antwortdetails
| Test | Observation | Inference | |
| (a) | C + 10cm\(^3\) of H\(_2\)O | C partially dissolves | C is a mixture of a soluble and insoluble salt. |
| b(i) | 2cm\(^3\) of filtrate + heat | Colourless and odourless gas in form of bubbles | CO\(_3\)\(^{2-}\) suspected |
| (ii) | residue + t.t + dil. HCl | No visible reaction | NO\(_3\)\(^-\), SO\(_4\)\(^{2-}\) suspected |
| (iii) | Solution from b(ii) + NaOH in drops and then in excess | White ppt which is dissolves in excess NaOH | Zn\(^{2+}\) , Pb\(^{2+}\) , Al\(^{3+}\) suspected |
| (iv) | Solution from b(ii) + NH\(_3\) in drops and then in excess | White ppt insoluble in excess NH\(_3\) | Pb\(^{2+}\) confirmed. |
Frage 4 Bericht
ALTERNATIVE PRACTICAL B
D contains 6.30 g of HNO\(_3\) in 500 cm\(_3\) of solution. E was prepared by dissolving 8.20 g of washing soda crystals ( Na\(_2\)CO\(_3\).xH\(_2\)O) in 250 cm\(^3\) of distilled water.
(a) Put D into the burette and titrate it against 20.0 or 25.0 cm\(^3\) portion of E using methyl orange as indicator.
Repeat the exercise to obtain concordance titre values.
Tabulate your results and calculate the average volume of the acid used.
(b) Write a balanced chemical equation for the reaction.
(c) From your results and the information given;
(i) calculate the concentration of E in moldm\(^{-3}\);
(ii) what is the value of x in Na\(_2\)CO\(_3\).xH\(_2\)O? [H= 1.0; C =12.0; O = 16.0; Na = 23.0; N = 14.0 ]
(a) Volume of pipette used = 25.00 cm\(^3\)
| Burette readings (cm\(^3\)) | Rough | 1st titre | 2nd titre |
| Final burette reading | 20.90 | 40.90 | 21.00 |
| Initial burette reading | 0.00 | 20.10 | 0.00 |
| Volume of acid, A used | 20.90 | 20.80 | 21.00 |
Volume of A used = \(\frac{Rough + 1st titre + 2nd titre}{3}\)
= \(\frac{20.90 + 20.80 + 21.00}{3}\)
= 20.90 cm\(^3\)
(b) 2HNO\(_3\)\(_{aq}\) + Na\(_2\)CO\(_3\).xH\(_2\)O\(_{aq}\) → 2NaNO\(_3\) + CO\(_2\) + (X+1)H\(_2\)0(l)
c(i) concentration of E in moldm\(^{-3}\)
Since we do not know the concentration of D in moldm\(^{-3}\), but in g per cm\(^3\), it will be difficult using the fomula \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\).
Thus, we must first convert the Conc. of D to moldm\(^{-3}\).
We were told from the question that D contains 6.30 g of HNO\(_3\) in 500 cm\(_3\),
So, if D 6.30 g of HNO\(_3\) was dissolved in 500 cm\(_3\)
then, X g of HNO\(_3\) will be dissolved in 1000 cm\(_3\)
X = \(\frac{{6.30}\times{1000}}{500}\) = 12.6 gdm\(^{-3}\)
Recall that Molar concentration (moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\)
Molar mass od HNO\(_3\) = 1 + 14 + (16 X 3) = 63 g/mol
concentration of D(moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\)
C\(_D\) = \(\frac{12.6}{63}\) = 0.20 moldm\(^{-3}\)
Now we can use the formula \(\frac{C_D V_D}{C_E V_E}\) = \(\frac{N_D}{N_E}\).
Given that
C\(_D\) = 0.20 moldm\(^{-3}\) N\(_D\) = 2
V\(_D\) = 20.90cm\(^3\) N\(_E\) = 1
C\(_E\) = ?
V\(_E\) = 25.0cm\(^3\)
C\(_B\) = \(\frac{C_D V_D N_E}{N_D V_E}\)
C\(_B\) = \(\frac{ 0.2 X 20.90 X 1}{25 X 2}\)
= 0.0836 moldm\(^{-3}\)
= 0.084 moldm\(^{-3}\)
(ii) To get the value of x in Na\(_2\)CO\(_3\).xH\(_2\)O
First, from the question, if 8.20 g of Na\(_2\)CO\(_3\).xH\(_2\)O was dissolved in 250 cm\(^3\) of distilled water.
then, Xg of Na\(_2\)CO\(_3\).xH\(_2\)O will be dissolved in 1000 cm\(^3\) of distilled water.
So, X = \(\frac{{8.20}\times{1000}}{250}\) = 32.8 gdm\(^{-3}\)
Mass concentration of E = 32.8 gdm\(^{-3}\)
Recall that concentration of E(moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\)
Then, Molar mass of E ( Na\(_2\)CO\(_3\).xH\(_2\)O ) = \(\frac{Mass concentration}{Molar concentration}\)
= \(\frac{32.8}{0.0836}\)
= 392.34 g/mol
But Molar mass of Na\(_2\)CO\(_3\).xH\(_2\)O = Na\(_2\)CO\(_3\) + x(H\(_2\)O)
392.34 = (23 x 2) + 12 + (16 x 3) + 18x
392.34 = 46 + 12 + 48 + 18x
392.34 = 106 + 18x
18x = 392.34 - 106
18x = 286.34
x = \(\frac{286.34}{18}\)
x = 15.9
x ≈ 16
Antwortdetails
(a) Volume of pipette used = 25.00 cm\(^3\)
| Burette readings (cm\(^3\)) | Rough | 1st titre | 2nd titre |
| Final burette reading | 20.90 | 40.90 | 21.00 |
| Initial burette reading | 0.00 | 20.10 | 0.00 |
| Volume of acid, A used | 20.90 | 20.80 | 21.00 |
Volume of A used = \(\frac{Rough + 1st titre + 2nd titre}{3}\)
= \(\frac{20.90 + 20.80 + 21.00}{3}\)
= 20.90 cm\(^3\)
(b) 2HNO\(_3\)\(_{aq}\) + Na\(_2\)CO\(_3\).xH\(_2\)O\(_{aq}\) → 2NaNO\(_3\) + CO\(_2\) + (X+1)H\(_2\)0(l)
c(i) concentration of E in moldm\(^{-3}\)
Since we do not know the concentration of D in moldm\(^{-3}\), but in g per cm\(^3\), it will be difficult using the fomula \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\).
Thus, we must first convert the Conc. of D to moldm\(^{-3}\).
We were told from the question that D contains 6.30 g of HNO\(_3\) in 500 cm\(_3\),
So, if D 6.30 g of HNO\(_3\) was dissolved in 500 cm\(_3\)
then, X g of HNO\(_3\) will be dissolved in 1000 cm\(_3\)
X = \(\frac{{6.30}\times{1000}}{500}\) = 12.6 gdm\(^{-3}\)
Recall that Molar concentration (moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\)
Molar mass od HNO\(_3\) = 1 + 14 + (16 X 3) = 63 g/mol
concentration of D(moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\)
C\(_D\) = \(\frac{12.6}{63}\) = 0.20 moldm\(^{-3}\)
Now we can use the formula \(\frac{C_D V_D}{C_E V_E}\) = \(\frac{N_D}{N_E}\).
Given that
C\(_D\) = 0.20 moldm\(^{-3}\) N\(_D\) = 2
V\(_D\) = 20.90cm\(^3\) N\(_E\) = 1
C\(_E\) = ?
V\(_E\) = 25.0cm\(^3\)
C\(_B\) = \(\frac{C_D V_D N_E}{N_D V_E}\)
C\(_B\) = \(\frac{ 0.2 X 20.90 X 1}{25 X 2}\)
= 0.0836 moldm\(^{-3}\)
= 0.084 moldm\(^{-3}\)
(ii) To get the value of x in Na\(_2\)CO\(_3\).xH\(_2\)O
First, from the question, if 8.20 g of Na\(_2\)CO\(_3\).xH\(_2\)O was dissolved in 250 cm\(^3\) of distilled water.
then, Xg of Na\(_2\)CO\(_3\).xH\(_2\)O will be dissolved in 1000 cm\(^3\) of distilled water.
So, X = \(\frac{{8.20}\times{1000}}{250}\) = 32.8 gdm\(^{-3}\)
Mass concentration of E = 32.8 gdm\(^{-3}\)
Recall that concentration of E(moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\)
Then, Molar mass of E ( Na\(_2\)CO\(_3\).xH\(_2\)O ) = \(\frac{Mass concentration}{Molar concentration}\)
= \(\frac{32.8}{0.0836}\)
= 392.34 g/mol
But Molar mass of Na\(_2\)CO\(_3\).xH\(_2\)O = Na\(_2\)CO\(_3\) + x(H\(_2\)O)
392.34 = (23 x 2) + 12 + (16 x 3) + 18x
392.34 = 46 + 12 + 48 + 18x
392.34 = 106 + 18x
18x = 392.34 - 106
18x = 286.34
x = \(\frac{286.34}{18}\)
x = 15.9
x ≈ 16
Frage 5 Bericht
(a) Mention two gases that are soluble in water.
(b) Name two apparatus that could be used to measure 9.50 cm\(^3\) of a solution accurately.
(c) What is the use of a fume cupboard in the laboratory?
(d) Name three personal protective equipment that are used in the laboratory.
(a) Gases that are soluble in water are:
- Ammonia
- Hydrogen chloride
- oxygen,
- carbon dioxide,
- Sulphur dioxide,
- chlorine, and
- hydrogen sulfide.
(b) Apparatus that could be used to measure volume of solutions accurately:
- volumetric flask,
- graduated cylinder,
- burette, or
- volumetric pipette.
(c) In a laboratory, fume cupboards are essential safety equipment used to capture and remove airborne hazardous or poisonous substances like gases, vapours, aerosols, and dust, protecting users from exposure to harmful chemicals.
(d) Personal protective equipment (PPE) used in the laboratory.
- Lab coats
- safety glasses
- respiratory protector
- Hand gloves
- Face shield
Antwortdetails
(a) Gases that are soluble in water are:
- Ammonia
- Hydrogen chloride
- oxygen,
- carbon dioxide,
- Sulphur dioxide,
- chlorine, and
- hydrogen sulfide.
(b) Apparatus that could be used to measure volume of solutions accurately:
- volumetric flask,
- graduated cylinder,
- burette, or
- volumetric pipette.
(c) In a laboratory, fume cupboards are essential safety equipment used to capture and remove airborne hazardous or poisonous substances like gases, vapours, aerosols, and dust, protecting users from exposure to harmful chemicals.
(d) Personal protective equipment (PPE) used in the laboratory.
- Lab coats
- safety glasses
- respiratory protector
- Hand gloves
- Face shield
Frage 6 Bericht
ALTERNATIVE TO PRACTICAL A
A is a solution of KMnO\(_4\) containing 0.020 moldm\(^{-3}\). B is a solution of Fe\(^{2+}\) obtained by dissolving 3.8 g of iron granules in 250 cm\(^3\) of solution.
(a)Put A in the burette. Pipette 20.0 cm\(^3\) or 25 cm\(^3\) of B into a conical flask and add 10 cm\(^3\) of H\(_2\)SO\(_4\). Titrate it with A.
Repeat the titration to obtain concordant titre values. Tabulate your result and calculate the average volume of A used.
MnO\(_4\)\(^-\)\(_{(aq)}\) + 5Fe\(^{2+}\)\(_{(aq)}\) + 8H\(^+\)\(_{(aq)}\) → Mn\(^{2+}\)\(_{(aq)}\) + Fe\(^{3+}\)\(_{(aq)}\) + 4H\(_2\)O\(_{(l)}\)
[ KMnO\(_4\) = 158.0, Fe = 56.0 ]
(b) From your results and information provided, calculate the:
(i) concentration of B in moldm\(^{-3}\)
(ii)concentration of B in gdm\(^{-3}\)
(iii) mass of Fe\(^{2+}\) in 250 cm\(^3\) of B.
(iv) percentage of Fe\(^{2+}\) in the granules.
(a) Volume of pipette used (Volume of B used) = 25.0 cm\(^3\)
| Burette Readings (cm\(^3\)) | Rough | 1st titre | 2nd titre |
| Final burette readings | 18.20 | 18.50 | 38.40 |
| Initial burette readings | 0.00 | 0.40 | 20.20 |
| Volume of A used (KMnO\(_4\)) | 18.20 | 18.10 | 18.20 |
Volume of A used = \(\frac{Rough + 1st titre + 2nd titre}{3}\)
= \(\frac{18.20 + 18.10 + 18.20}{3}\)
= 18.166
V\(_A\) = 18.17 cm\(^3\)
(i) Concentration of B in moldm\(^{-3}\)
From the balanced equation : MnO\(_4\)\(^-\)\(_{(aq)}\) + 5Fe\(^{2+}\)\(_{(aq)}\) + 8H\(^+\)\(_{(aq)}\) → Mn\(^{2+}\)\(_{(aq)}\) + Fe\(^{3+}\)\(_{(aq)}\) + 4H\(_2\)O\(_{(l)}\)
C\(_A\) = 0.020 moldm\(^{-3}\) N\(_A\) = 1
V\(_A\) = 18.17cm\(^3\) N\(_B\) = 5
C\(_B\) = ?
V\(_B\) = 25.0cm\(^3\)
Using \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\)
C\(_B\) = \(\frac{C_A V_A N_B}{N_A V_B}\)
C\(_B\) = \(\frac{ 0.02 X 18.17 X 5}{25 X 1}\)
C\(_B\) = 0.07268
C\(_B\) = 0.070 moldm\(^{-3}\)
(ii) Concentration of B in gdm\(^{-3}\)
If 3.8 g of Fe granules was dissolved in 250 cm\(^3\) of B
Xg of Fe granules will be dissolved in 1000 cm\(^3\) of B
X = \(\frac{ 3.8 X 1000}{250}\) = 15.2 gdm\(^{-3}\)
Concentration of B in gdm\(^{-3}\) will be 15.2 gdm\(^{-3}\)
(iii) Mass of Fe\(^{2+}\) in 250cm\(^3\) of B.
To get the mass, we'll use the formula n = \(\frac{mass}{Molar mass}\)
However, since do not know the number of moles, n , hence we'll need to obtain n, using n = \(\frac{CV}{1000}\) since we have C and V
So, n = \(\frac{{0.070}\times{250}}{1000}\) = 0.025 mole
Now, we can apply n = \(\frac{mass}{Molar mass}\) to obtain the mass, given that the Molar mass of Fe was given to be 56
n = \(\frac{mass}{Molar mass}\)
m = n x M = 0.018 x 56 = 1.02 g
Therefore, 1.02 g of Fe\(^{2+}\) in 250 cm\(^3\) of B.
(iv) Percentage of Fe\(^{2+}\) in the granules = \(\frac{Mass of Fe^{2+}}{Total mass of Fe granules}\times 100\)
= \(\frac{1.02}{3.8}\times 100\)
= 26.89%
= 26.9%
Antwortdetails
(a) Volume of pipette used (Volume of B used) = 25.0 cm\(^3\)
| Burette Readings (cm\(^3\)) | Rough | 1st titre | 2nd titre |
| Final burette readings | 18.20 | 18.50 | 38.40 |
| Initial burette readings | 0.00 | 0.40 | 20.20 |
| Volume of A used (KMnO\(_4\)) | 18.20 | 18.10 | 18.20 |
Volume of A used = \(\frac{Rough + 1st titre + 2nd titre}{3}\)
= \(\frac{18.20 + 18.10 + 18.20}{3}\)
= 18.166
V\(_A\) = 18.17 cm\(^3\)
(i) Concentration of B in moldm\(^{-3}\)
From the balanced equation : MnO\(_4\)\(^-\)\(_{(aq)}\) + 5Fe\(^{2+}\)\(_{(aq)}\) + 8H\(^+\)\(_{(aq)}\) → Mn\(^{2+}\)\(_{(aq)}\) + Fe\(^{3+}\)\(_{(aq)}\) + 4H\(_2\)O\(_{(l)}\)
C\(_A\) = 0.020 moldm\(^{-3}\) N\(_A\) = 1
V\(_A\) = 18.17cm\(^3\) N\(_B\) = 5
C\(_B\) = ?
V\(_B\) = 25.0cm\(^3\)
Using \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\)
C\(_B\) = \(\frac{C_A V_A N_B}{N_A V_B}\)
C\(_B\) = \(\frac{ 0.02 X 18.17 X 5}{25 X 1}\)
C\(_B\) = 0.07268
C\(_B\) = 0.070 moldm\(^{-3}\)
(ii) Concentration of B in gdm\(^{-3}\)
If 3.8 g of Fe granules was dissolved in 250 cm\(^3\) of B
Xg of Fe granules will be dissolved in 1000 cm\(^3\) of B
X = \(\frac{ 3.8 X 1000}{250}\) = 15.2 gdm\(^{-3}\)
Concentration of B in gdm\(^{-3}\) will be 15.2 gdm\(^{-3}\)
(iii) Mass of Fe\(^{2+}\) in 250cm\(^3\) of B.
To get the mass, we'll use the formula n = \(\frac{mass}{Molar mass}\)
However, since do not know the number of moles, n , hence we'll need to obtain n, using n = \(\frac{CV}{1000}\) since we have C and V
So, n = \(\frac{{0.070}\times{250}}{1000}\) = 0.025 mole
Now, we can apply n = \(\frac{mass}{Molar mass}\) to obtain the mass, given that the Molar mass of Fe was given to be 56
n = \(\frac{mass}{Molar mass}\)
m = n x M = 0.018 x 56 = 1.02 g
Therefore, 1.02 g of Fe\(^{2+}\) in 250 cm\(^3\) of B.
(iv) Percentage of Fe\(^{2+}\) in the granules = \(\frac{Mass of Fe^{2+}}{Total mass of Fe granules}\times 100\)
= \(\frac{1.02}{3.8}\times 100\)
= 26.89%
= 26.9%
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