14.0 g of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273 K. Which of the following statements is incorrect? [Given : Molar mass in mathrmg \ mol^-1$\mathrm{g \ mol^{-1}}$ of Ca–40, Cl–35.5, H–1]

### Related Formula mathrmCa_(mathrms) + 2mathrmHCl_(mathrmg) longrightarrow mathrmCaCl_2(mathrms) + mathrmH_2(mathrmg)$$\mathrm{Ca}_{(\mathrm{s})} + 2\mathrm{HCl}_{(\mathrm{g})} \longrightarrow \mathrm{CaCl}_{2(\mathrm{s})} + \mathrm{H}_{2(\mathrm{g})}$$ ### Core Logic Number of moles of Calcium (n_Ca$n_{Ca}$) = fractextGiven masstextMolar mass = frac14.040 = 0.35$\frac{\text{Given mass}}{\text{Molar mass}} = \frac{14.0}{40} = 0.35$ mol. Since HCl is in excess, Calcium is the limiting reagent. From stoichiometry, 1 mole of Ca produces 1 mole of H_2$H_2$ gas and 1 mole of CaCl_2$CaCl_2$. Moles of H_2$H_2$ evolved = 0.35 mol. Volume of H_2$H_2$ at STP (1 atm, 273 K) = 0.35 times 22.4$0.35 \times 22.4$ L = 7.84$7.84$ L. Mass of CaCl_2$CaCl_2$ produced = 0.35 times textMolar mass of CaCl_2$0.35 \times \text{Molar mass of } CaCl_2$ = 0.35 times (40 + 71) = 0.35 times 111 = 38.85$0.35 \times (40 + 71) = 0.35 \times 111 = 38.85$ g. Option (3) states 33.3 g of CaCl_2$CaCl_2$ is produced, which is incorrect. ### Step 1: Final Conclusion Statement (3) is incorrect. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry