Question

1. determine the theoretical yield (in g) of hcl if 65.0 g of bcl₃ and 27.5 g of h₂o are reacted according to the following balanced reaction. the molar mass of bcl₃ is 117.16 g mol⁻¹.

bcl₃(g) + 3h₂o(l) → h₃bo₃(s) + 3hcl(g)
a) 167 g hcl
b) 60.7 g hcl
c) 55.6 g hcl
d) 20.2 g hcl
e) 6.74 g hcl

Explanation:

Step1: Calculate moles of $BCl_3$

$n_{BCl_3}=\frac{m_{BCl_3}}{M_{BCl_3}}=\frac{65.0\ g}{117.16\ g/mol}\approx0.555\ mol$

Step2: Calculate moles of $H_2O$

The molar mass of $H_2O$ is $18.02\ g/mol$. So $n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{27.5\ g}{18.02\ g/mol}\approx1.53\ mol$

Step3: Determine the limiting reactant

From the balanced - equation $BCl_3(g)+3H_2O(l)
ightarrow H_3BO_3(s)+3HCl(g)$, the mole ratio of $BCl_3$ to $H_2O$ is $1:3$.
For $0.555\ mol$ of $BCl_3$, the moles of $H_2O$ required is $n_{H_2O - required}=3\times n_{BCl_3}=3\times0.555\ mol = 1.665\ mol$. But we have only $1.53\ mol$ of $H_2O$.
For $1.53\ mol$ of $H_2O$, the moles of $BCl_3$ required is $n_{BCl_3 - required}=\frac{1}{3}\times n_{H_2O}=\frac{1}{3}\times1.53\ mol = 0.51\ mol$. Since $0.51\ mol<0.555\ mol$, $H_2O$ is the limiting reactant.

Step4: Calculate moles of $HCl$ produced

From the balanced - equation, the mole ratio of $H_2O$ to $HCl$ is $3:3 = 1:1$. So the moles of $HCl$ produced is equal to the moles of $H_2O$ reacted. $n_{HCl}=n_{H_2O}=1.53\ mol$

Step5: Calculate the mass of $HCl$ produced

The molar mass of $HCl$ is $36.46\ g/mol$. So $m_{HCl}=n_{HCl}\times M_{HCl}=1.53\ mol\times36.46\ g/mol\approx55.6\ g$

Answer:

C. $55.6\ g\ HCl$