What weight of ${CaCO}_{3}$ must be decomposed to produce the sufficient quantity of carbon dioxide to convert 21.2 kg of ${Na}_{2} {CO}_{8}$ completely in to ${NaHCO}_{3}$ , [Atomic mass $Na = 23, Ca = 40$ ]
$\begin{array}{l} {CaCO}_{3} ⟶ CaO + {CO}_{2} \\ {Na}_{2} {CO}_{3} + {CO}_{2} + H_{2} O ⟶ 2 {NaHCO}_{3} \end{array}$

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100 Kg

20 Kg

120 Kg

30 Kg

answer is A.

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Detailed Solution

${CaCO}_{3} \to CaO + {CO}_{2}$

$100 - 56 \to 44$

$\begin{array}{l} {Na}_{2} {CO}_{3} + {CO}_{2} + H_{2} O \to 2 {NaHCO}_{3} \\ 106 + 18 + 44 \to 2 \times 84 \end{array}$

Now you have 20 kg so $21200 \times \frac{44}{106}$ .

$= 8.8 kg$ of ${CO}_{2}$ needed

From eq.1 $88000 \times \frac{100}{44}$

$= 20 kg$ of ${CaCO}_{3}$ is needed.

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