Bright Tutorials Bright Tutorials
ICSE Class 10 Chemistry Question 8 of 28

Stoichiometry — Percentage Composition — Empirical & Molecular Formula — Chemical Equations — Question 16

Back to all questions
16
Question

Question 1(2004)

The reaction of potassium permanganate with acidified iron (II) sulphate is given below :

2KMnO4 + 10FeSO4 + 8H2SO4 ⟶ K2SO4 + 2MnSO4 + 5 Fe2(SO4)3 + 8H2O.

If 15.8 g. of potassium permanganate was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction.

[K = 39, Mn = 55, Fe = 56, S = 32, O = 16]

Answer

2KMnO4+10FeSO4+8H2SO42[39+5510[56+32+4(16)]+4(16)]=2[39+10[56+55+64]32+64]=316 g=1520 g\begin{matrix} 2\text{KMnO}_4 & + 10 \text{FeSO}_4 & + & 8\text{H}_2\text{SO}_4 \\ 2[39 + 55 & 10[56+ 32 \\ + 4(16)] & + 4(16)] \\ = 2[39 + & 10[56 + \\ 55 + 64] & 32 + 64] \\ = 316 \text{ g} & = 1520\text{ g} \\ \end{matrix}

⟶K2SO4+2MnSO4+5Fe2(SO4)3+8H2O\longrightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 5 \text{Fe}_2(\text{SO}_4)_3 + 8\text{H}_2\text{O}

316 g of potassium permanganate was used with 1520 g of iron (II) sulphate

∴ 15.8 g. of potassium permanganate will be used with 1520316\dfrac{1520}{316} x 15.8 = 76 g of iron (II) sulphate

Hence, 76 g of iron (II) sulphate was used.