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Standard XII

Chemistry

Stoichiometric Calculations

6.2 g of a sa...

Question

$6.2$ g of a sample containing $N a_{2} C O_{3}, N a H C O_{3}$ and non-volatile inert impurity on gentle heating loses $5 %$ of its weight due to reaction as shown below:
$2 N a H C O_{3} \to N a_{2} C O_{3} + H_{2} O + C O_{2}$
Residue is dissolved in water and formed $100$ mL solution and its $10$ mL portion requires $7.5$ mL of $0.2 M$ aqueous solution of $B a C l_{2}$ for complete precipitation of carbonates. Determine the weight (in g) of $N a_{2} C O_{3}$ in the original sample (as nearest integer).

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Solution

Loss in weight due to heating $= 6.2 \times \frac{5}{100} = 0.31$ g
$2 N a H C O_{3} \to N a_{2} C O_{3} + C O_{2} + H_{2} O$
$2 \times 84 g 44 + 18 g$ loss
1 mole $N a H C O_{3} = 31$ g loss due to heating.
$0.31$ g loss is from $0.01$ mol of $N a H C O_{3}$ .
Moles of $N a_{2} C O_{3}$ produced $= \frac{0.01}{2} = 0.005$
Total moles of carbonate reacted with $B a C l_{2} = (7.5 \times 0.2 \times \frac{100}{10}) \times 10^{- 3} = 0.015$
Moles of carbonates in original sample $= 0.015 - 0.005 = 0.001$
Mass of $N a_{2} C O_{3}$ in original sample $= 0.01 \times 106 = 1.06$ g
Hence, the answer is $1$ .

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