Question

Hydrogen sulfide burns form sulfur dioxide:

2H2S(g) +3O2(g) → 2SO2(g) + 2H2O(g) ΔH= -1036 kJ


Calculate the enthalpy change when burning 26.7 g of hydrogen sulfide in kJ

pls help

Answer 1

Answer: 404.04 kJ.

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
`H_2S`

`\text{Number of moles}=(26.7g)/(34.1g/mol)=0.78moles`

`2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)`
`\Delta H=-1036kJ`

According to stoichiometry :

2 moles of
`H_2S` on burning produces = 1036 kJ

Thus 0.78 moles of
`H_2S` on burning produces =
`(1036kJ)/(2)* 0.78=404.04`

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.