Determine the theoretical yield of HCl if 60.0 g of BCl₃and 37.5 g of H₂O are reacted according to the following balanced reaction.A possibly useful molar mass is BCl₃ = 117.16 g/mol. BCl₃(g)+ 3 H₂O(l)→ H₃BO₃(s)+ 3 HCl(g) A) 75.9 g HCl B) 132 g HCl C) 187 g HCl D) 56.0 g HCl E) 25.3 g HCl

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Quizplus · Accepted Answer

First, calculate the moles of BCl3 and H2O. For BCl3: $60.0 \, \text{g} \div 117.16 \, \text{g/mol} = 0.512 \, \text{mol}$. For H2O: $37.5 \, \text{g} \div 18.015 \, \text{g/mol} = 2.08 \, \text{mol}$. The reaction shows BCl3 reacts with 3 moles of H2O to produce 3 moles of HCl. BCl3 is the limiting reagent, so it will produce $0.512 \, \text{mol} \times 3 = 1.536 \, \text{mol}$ of HCl. The mass of HCl produced is $1.536 \, \text{mol} \times 36.46 \, \text{g/mol} = 56.0 \, \text{g}$.

Determine the Theoretical Yield of HCl If 60