Question

Consider the following reaction: 2CH3OH(g)  2CH4(g) + O2(g) ΔH = +252.8 kJ a) Calculate the amount of heat transferred when 24.0 g of CH3OH(g) is decomposed by this reaction at constant pressure. b) For a given sample of CH3OH, the enthalpy change during the reaction is 82.1 kJ. How many grams of methane gas are produced?

Answer 1

Answer:

For a: The amount of heat transferred for the given amount of methanol is 94.6736 kJ.

For b: The mass of methane gas produced will be 10.384 g.

Explanation:

For the given chemical reaction:

$2CH_3OH(g)\rightarrow 2CH_4(g)+O_2(g);\Delta H=+252.8kJ$

• For a:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of methanol = 24.0 g

Molar mass of methanol = 32.04 g/mol

Putting values in above equation, we get:

$\text{Moles of methanol}=\frac{24.0g}{32.04g/mol}=0.749mol$

By Stoichiometry of the reaction:

For every 2 moles of methanol, the amount of heat transferred is +252.8 kJ.

So, for every 0.749 moles of methanol, the amount of heat transferred will be = $\frac{252.8}{2}\times 0.749=94.6736kJ$

Hence, the amount of heat transferred for the given amount of methanol is 94.6736 kJ.

• For b:

By Stoichiometry of the reaction:

252.8 kJ of energy is absorbed when 2 moles of methane gas is produced.

So, 82.1 kJ of energy will be absorbed when = $\frac{2}{252.8}\times 82.1=0.649mol$ of methane gas is produced.

Now, calculating the mass of methane gas from equation 1, we get:

Molar mass of methane gas = 16 g/mol

Moles of methane gas = 0.649 moles

Putting values in equation 1, we get:

$0.649mol=\frac{\text{Mass of methane gas}}{16g/mol}\\\\\text{Mass of methane}=10.384g$

Hence, the mass of methane gas produced will be 10.384 g.