Consider the unbalanced chemical equation below: $$\mathrm{CaSiO}_{3}(s)+\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}( )$$ Suppose a 32.9 -g sample of \(\mathrm{CaSiO}_{3}\) is reacted with 31.8 \(\mathrm{L}\) of \(\mathrm{HF}\) at \(27.0^{\circ} \mathrm{C}\) and 1.00 \(\mathrm{atm}\) . Assuming the reaction goes to completion, calculate the mass of the \(\mathrm{SiF}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced in the reaction.

#tag_title#Step 2: Determine moles of reactants and the limiting reactant#tag_content# Moles of CaSiO₃ = $\frac{32.9\,\text{g}}{116.16\,\text{g/mol}} = 0.283\,\text{mol}$ Moles of HF = $\frac{31.8\,\text{L}\times 1.00\,\text{atm}}{0.08206\,\text{L}\cdot\text{atm/mol}\cdot\text{K}\times(27.0^{\circ}C+273.15)} = 1.32\,\text{mol}$ Limiting reactant: $\frac{1.32\,\text{mol}\,\text{HF}}{4} = 0.330\,\text{mol} \gt 0.283\,\text{mol}\,\text{CaSiO}_{3}$, so CaSiO₃ is the limiting reactant. #tag_title#Step 3: Calculate mass of products formed#tag_content# Moles of SiF₄ = moles of CaSiO₃ = 0.283 mol Moles of H₂O = $2\times0.283\,\text{mol} = 0.566\,\text{mol}$ Mass of SiF₄ = $0.283\,\text{mol} \times 104.08\,\text{g/mol} = 29.4\,\text{g}$ Mass of H₂O = $0.566\,\text{mol} \times 18.015\,\text{g/mol} = 10.2\,\text{g}$ So, 29.4 g of SiF₄ and 10.2 g of H₂O are produced in the reaction.