Q20 / 180 · COMEDK 2012 · Chemistry

The volume of $\mathrm{CO}_{2}\left(\mathrm{in} \mathrm{cm}^{3}\right)$ liberated at STP, when $1.06 \mathrm{~g}$ of anhydrous sodium carbonate is treated with excess of dilute $\mathrm{HCl}$ is [atomic mass of $\mathrm{Na}=23]$

  • A
    112
  • Correct answer
    B
    224
  • C
    56
  • D
    2240

Solution

$\mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{HCl}$ (dil.) $\longrightarrow 2 \mathrm{NaCl}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$ $1 \mathrm{~mol} \quad$ (dil.) $1 \mathrm{~mol}$ $=1.06 \mathrm{~g} \quad=22400 \mathrm{~cm}^{3}$ $1.06 \mathrm{~g} \mathrm{Na}_{2} \mathrm{CO}_{3}$ gives $=22400 \mathrm{~cm}^{3}$ of CO$_{2}$ $\therefore 1.06$ g gives $=\frac{1.06 \times 22400}{106}=224 \mathrm{~cm}^{3}$ of $\mathrm{CO}_{2}$

Hence, the correct answer is option: 'B'.

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