A $2.10$ g mixture of $N a H C O_{3}$ and $K C l O_{3}$ requires $100$ mL of $0.1 N$ $H C l$ for complete reaction. Calculate the amount of residue (as nearest integer) that would be obtained on strongly heating $2.20$ g of the same mixture.

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Solution

Only $N a H C O_{3}$ reacts with $0.1 N$ $H C l$ .
Let $x$ g be the weight of $N a H C O_{3}$ and $(2.1 - x)$ g be the weight of $K C l O_{3}$ .
Milliequivalents of $N a H C O_{3} = 100 \times 0.1 = 10$
Weight of $N a H C O_{3} = 10 \times 10^{- 3} \times 84 = 0.84 g$
Weight of $K C l O_{3} = 2.1 - 0.84 = 1.26 g$
$2 N a H C O_{3} Δ \to N a_{2} C O_{3} + H_{2} O ↑ + C O_{2} ↑$
$2$ mol of $N a H C O_{3} = 2 \times 82$ g
$N a_{2} C O_{3} = 106$ g
$2 K C l O_{3} Δ \to 2 K C l + 3 O_{2} ↑$
The residue will have $N a_{2} C O_{3}$ and $K C l$ .
$K C l O_{3} = 122.5$ g
$K C l = 74.5$ g
Weight of residue from $2.1$ g of mixture $= \frac{53}{84} \times 0.84 + \frac{74.5}{122.5} \times 1.26$ $= 0.53 + 0.766 = 1.296$ g
Weight of residue from $2.2$ g of the mixture $= 0.84 + 1.26 = 2.20$ g $= \frac{1.296}{2.1} \times 2.2 = 1.358$ g
Hence, the answer to the nearest integer is $1$ .

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