Question

Hydrogen sulfide burns form sulfur dioxide:

Answer 1

Answer: 404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

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

moles of $H_2S$

$\text{Number of moles}=\frac{26.7g}{34.1g/mol}=0.78moles$

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

According to stoichiometry :

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

Thus 0.78 moles of $H_2S$ on burning produces =$\frac{1036kJ}{2}\times 0.78=404.04$

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