Question

chloroform, $chcl_3$, reacts with chlorine, $cl_2$, to form carbon tetrachloride, $ccl_4$, and hydrogen chloride, $hcl$. in an experiment 25 grams of chloroform and 25 grams of chlorine were mixed. which is the limiting reagent?
$chcl_3 + cl_2 \to ccl_4 + hcl$
a $hcl$
b $chcl_3$
c $ccl_4$
d $cl_2$

Explanation:

Step1: Balance the chemical equation

$$\mathrm{CHCl_3 + Cl_2 ightarrow CCl_4 + HCl}$$

Balanced equation: $$\mathrm{CHCl_3 + Cl_2 = CCl_4 + HCl}$$
The mole ratio of $\mathrm{CHCl_3}$ to $\mathrm{Cl_2}$ is $1:1$.

Step2: Calculate the molar mass

• Molar mass of $\mathrm{CHCl_3}$: $M_{CHCl_3}=12 + 1+3\times35.5=119.5\ \text{g/mol}$
• Molar mass of $\mathrm{Cl_2}$: $M_{Cl_2}=2\times35.5 = 71\ \text{g/mol}$

Step3: Calculate the number of moles

• Moles of $\mathrm{CHCl_3}$: $n_{CHCl_3}=\frac{m_{CHCl_3}}{M_{CHCl_3}}=\frac{25\ \text{g}}{119.5\ \text{g/mol}}\approx0.21\ \text{mol}$
• Moles of $\mathrm{Cl_2}$: $n_{Cl_2}=\frac{m_{Cl_2}}{M_{Cl_2}}=\frac{25\ \text{g}}{71\ \text{g/mol}}\approx0.35\ \text{mol}$

Since the mole ratio of $\mathrm{CHCl_3}$ to $\mathrm{Cl_2}$ is $1:1$ in the balanced reaction, and $n_{CHCl_3}(0.21\ \text{mol})

Answer:

B. $\mathrm{CHCl_3}$