Question

Methane (CH4) reacts with Cl2 to yield CCl4 and HCl by the following reaction equation: CH4 + 4 Cl2 → CCl4 + 4HCl. What is the ΔH of the reaction if 51.3 g of CH4 reacts with excess Cl2 to yield 1387.6 kJ?

Answer 1

Answer: The ΔH of the reaction if 51.3 g of $CH_4$ reacts with excess $Cl_2$ to yield 1387.6 kJ is 432.27kJ

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $CH_4$

$\text{Number of moles of methane}=\frac{51.3g}{16g/mol}=3.21moles$

$CH_4+4Cl_2\rightarrow CCl_4+4HCl$   $\Delta H=?$

As $Cl_2$ is present in excess, $CH_4$ is the limiting reagent as it limits the formation of product.

If 3.21 moles of methane releases heat = 1387.6 kJ

Thus 1 mole of methane release=$\frac{1387.6}{3.21}\times 1=432.27kJ$