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

Step-by-step 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}=(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=
`(1387.6)/(3.21)* 1=432.27kJ`