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Stoichiometry

1.249 g of a ...

Question

$1.249$ g of a sample of pure $B a C O_{3}$ and impure $C a C O_{3}$ containing some $C a O$ was treated with dilute $H C l$ and it evolved $168$ mL of $C O_{2}$ at NTP. From this solution, $B a C r O_{4}$ was precipitated, filtered and washed. The dried precipitate was dissolved in dilute sulphuric acid and diluted to $100$ mL. $10$ mL of this solution when treated with $K I$ solution liberated iodine, which required exactly $20$ mL of $0.05 N$ $N a_{2} S_{2} O_{3}$ . Calculate the percentage of $C a O$ in the sample.

14.09 %

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7.04

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70.4

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1.409 %

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Solution

The correct option is A $14.09 %$
Let the mass of $B a C O_{3}$ , $C a C O_{3}$ and $C a O$ be $a$ , $b$ and $c$ respectively.
$∴ a + b + c = 1.249 . . . (i)$
Meq. of $B a C O_{3}$ + Meq. of $C a C O_{3} =$ Meq. of $C O_{2}$
$\frac{a}{(\frac{197}{2})} \times 1000 + \frac{b}{(\frac{100}{2})} \times 1000 = \frac{168 \times 44}{22400 \times 22} \times 1000$
$∴ 200 a + 394 b = 295.5 . . . (i i)$
Also, for the redox changes,
$B a C O_{3} ⟶ B a C r O_{4}$
$3 e + C r^{6 +} ⟶ C r^{3 +}$
$2 I^{-} ⟶ I_{2} + 2 e$
Meq. of $B a C O_{3} =$ Meq. of $B a C r O_{4} =$ Meq. of $I_{2}$
$∴ \frac{a}{(\frac{197}{3})} \times = 20 \times 0.005 \times \frac{100}{10}$
$∴ a = 0.656$ g $. . . (i i i)$
(Note : Equivalent mass of $B a C r O_{4}$ is $\frac{M}{3}$ and thus, for $B a C O_{3}$ it should be $\frac{M}{3}$ )
By Eqs. $(i i)$ and $(i i i)$ , $b = 0.417$ g
$∴$ from Eq. $(i)$ , $0.656 + 0.417 + c = 1.249$
$∴ c = 0.176$
Percentage of $C a O = \frac{0.176 \times 100}{1.249} = 14.09 %$

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