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Hydrides of Nitrogen

A mixture of ...

Question

A mixture of iron (II) and lead (II) carbonates was calcined. As a result of this, an amount of $0.9$ mol of $C O_{2}$ was evolved. The mixture of metal oxides obtained by decomposing the above carbonates was treated with sulphuric acid which resulted in a mass of $151.6$ g of a sulphate as precipitate. Determine the composition of the starting mixture of carbonates in mass percentage.

Mass of

F e C O_{3} = 25.79 %

, Mass of

P b C O_{3} = 74.21 %

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Mass of

F e C O_{3} = 15.79 %

, Mass of

P b C O_{3} = 84.21 %

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Mass of

F e C O_{3} = 35.79 %

, Mass of

P b C O_{3} = 64.21 %

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None of these

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Solution

The correct option is A Mass of $F e C O_{3} = 25.79 %$ , Mass of $P b C O_{3} = 74.21 %$
Calcination of the metal carbonates:
(a) $F e C O_{3} \to F e O + C O_{2}$
$P b C O_{3} \to P b O + C O_{2}$
Treatment of metal oxides with sulphuric acid:
(b) $F e O + H_{2} S O_{4} \to F e S O_{4} + H_{2} O$
$P b O + H_{2} S O_{4} \to P b S O_{4} ↓ + H_{2} O$
Only $P b S O_{4}$ was precipitated out which weigh $= 151.6$ g.
Let $x$ mol of $F e C O_{3}$ and $y$ mol of $P b C O_{3}$ are present in the mixture.
Therefore, moles of $C O_{2} = x + y = 0.9$
Moles of $P b S O_{4}$ = Moles of $P b O$ = Moles of $P b C O_{3} = y$
$y = \frac{151.6}{303 (M o l e c u l a r w e i g h t o f P b S O_{4})} = 0.5$
$∴ x = 0.4$
Mass of $F e C O_{3} = 0.4 \times 116 = 46.9 g = 25.79 %$
Mass of $P b C O_{3} = 0.5 \times 267 = 133.5 g = 74.21 %$

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