Question

Sodium carbonate of 92% purity is used in
the reation Na,Co, + CaCl, → Caco, +
2 NaCl. The number of grams of Na,CO2
required to yield 1gm of CaCO,
​

Answer 1

1 mole of Na2co3 is react with 1 mole of Caco3 so ,

if 100g of caco3 = 106 g of na2co3 then 1g of caco3 = 106*1/100

so answer is 1.06g

Answer 2

Answer: 1.152 g

Explanation:

$Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of calcium carbonate}=\frac{1g}{100g/mol}=0.01moles$

According to stoichiometry:

1 mole of $CaCO_3$ is produced from 1 mole of $Na_2CO_3$

Thus 0.01 moles of $CaCO_3$ are produced from=$\frac{1}{1}\times 0.010=0.010moles$ of $Na_2CO_3$

But as percent purity of $Na_2CO_3$  is 92% ,

moles of $Na_2CO_3=\frac{100}{92}\times 0.010=0.011moles$

Mass of $Na_2CO_3=mass\times {\text {Molar mass}}=0.011\times 106=1.152g$

The number of grams of  $Na_2CO_3$ required to yield 1 gm of $CaCO_3$ is 1.152 g.