Question

The volume of $CO_2$ (in $cm^3$) liberated at S.T.P. when 1.06 g of anhydrous sodium carbonate is treated with excess of dilute HCl is [atomic mass of Na= 23]

• A. 112
• B. 224
• C. 56
• D. 2240

Answer 1

Answer: (B)

224

Answer 2

${ Na_2CO_3 + 2HCl -> 2NaCl +CO_2 + H_2O}$
106 g $22400 cm^3$
1.06 g $\frac{22400}{106} \times 1.06 \, cm^2$
= $224 \, cm^3$