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

Step-by-step 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}=(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 =
`(431.8)/(0.771)* 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