Question

Volume of 0.1 M HCl required to react completely with 1 g equimolar mixture of $Na_2CO_3$ and $NaHCO_3$ is

• A. 100 mL
• B. 200 mL
• C. 157.9 mL
• D. 159.7 mL

Answer 1

Answer: (C)

157.9 mL

Answer 2

${NaCO_3 + 2HCl -> 2NaCl + H_2O + CO_2}$ ${ NaHCO_3 + HCl -> NaCl + H_2O + CO_2}$ If the mixture contains 1 mol of each, HCl required is (2 + 1) = 3 mol . Molar mass of $Na_2CO_3$ = 106 g mol$^{-1}$ . Molar mass of $NaHCO_3$ = 84 g mol$^{-1}$ . Thus, 190 g (106 g + 84 g) of mixture require HCI= 3 mol , 1 g of mixture require HCl = $\frac{3}{190}$ mol $\therefore$ Vol of 0.1 M HCl which contains (3/190) mol = $\frac{1000 \times 3}{190 \times 0.1}$ = 157.89 mL.