UPSEE Solved Model Paper 7

Given: Mass of $\mathrm{CaCO}_3$ = 10 g Volume of HCl = 1 L Molarity of HCl = 0.1 M We have to find the volume of CO2 produced. Chemical reaction of calcium carbonate: $\mathrm{CaCO}_3$ with HCl is given as： $\mathrm{CaCO}_3$ + 2HCl → $\mathrm{CaCl}_2 + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}$ ... (i) Molecular mas of $\mathrm{CaCO}_3$ = 40 + 12 + 3 × 16 = 100 g Now, number of moles is given as : Number of moles = $\frac{\text{Mass}}{\text{Molecular mass}}$ ... (i) Thus, no. of moles of $\mathrm{CaCO}_3$ = $\frac{10}{100}$ = 0.1 moles Now, molarity is given as Molarity = $\frac{\text{No. of moles solute}}{\text{Volume of solution (in mL)}}$ ... (ii) Substituting the values of molarity of HCl and volume of HCl in equation (i), we get 0.1 = $\frac{\text{No. of moles}}{1}$ Thus, number of moles of HCl = 0.1 From reaction (i), 1 mole of $\mathrm{CaCO}_3$ reacts with 2 moles of HCl to produce 1 mole of $\mathrm{CO}_2$ Moles of HCl required is double the moles of $\mathrm{CaCO}_3$. But moles of HCl is equal to moles of $\mathrm{CaCO}_3$, i.e., 0.1. Thus, HCl is the limiting reagent From reaction (i) 2 moles of HCl produce 1 mole of $\mathrm{CO}_2$ Thus, 0.1 mole of HCl will produce = $\frac{1}{2} \times 0.1$ = 0.05 moles of $\mathrm{CO}_2$ 1 mole of $\mathrm{CO}_2$ = 22.4 L of volume = 22.4 × $10^3 \text{ cm}^3$ (Since 1 L = 1000 $\text{cm}^3$) Thus, 0.05 moles of $\mathrm{CO}_2$ = 0.05 × 22.4 × $10^3$ = 1120 $\text{cm}^3$ Hence, option 'A' is correct