Question

Hydrosulfuric acid (H2S) undergoes combustion to yield sulfur dioxide and water by the following reaction equation: 2 H2S + 3 O2 → 2 SO2 + 2 H2O

Answer 1

The question is incomplete, here is the complete question:

Hydrosulfuric acid $(H_2S)$ undergoes combustion to yield sulfur dioxide and water by the following reaction equation:

$2H_2S+3O_2\rightarrow 2SO_2+2H_2O$

What is the $\Delta H$ of the reaction if 26.2 g of $H_2S$ reacts with excess $O_2$ to yield 431.8 kJ?

Answer: The $\Delta H$ of the reaction is -1120.10 kJ

Explanation:

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

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

Given mass of hydrogen sulfide = 26.2 g

Molar mass of hydrogen sulfide = 34 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrogen sulfide}=\frac{26.2g}{34g/mol}=0.771mol$

We are given:

Amount of heat released = 431.8 kJ

By Stoichiometry of the reaction:

When 0.771 moles of hydrogen sulfide is reacted, the amount of heat released is 431.8 kJ

So, when 2 moles of hydrogen sulfide will react, the amount of heat released will be = $\frac{431.8}{0.771}\times 2=1120.10kJ$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

Hence, the $\Delta H$ of the reaction is -1120.10 kJ