Question

The burning of 48.7 g of zns in the presence of oxygen gives 220.0 kj of heat. what is h for the reaction as written below?

Answer 1

The ΔH of the reaction when 48.7 grams of ZnS reacts with oxygen is -55.23 kJ.

Step-by-step explanation:

`2ZnS+3O_2\rightarrow 2ZnO+2SO_2`

Amount of ZnS = 48.7 grams

Molecular mass of ZnS = 97.474 g/mol

`\Delta H=-220 kJ`

Moles of ZnS.:

`\text{ Moles of ZnS}=\frac{\text{ Mass of ZnS}}{\text{ Molar mass of ZnS}}=(48.7g)/(97.474g/mole)=0.5021 mol`

According to reaction, 2 moles of ZnS gives energy = -220 kJ

So, 0.5021 moles of ZnS gives energy :

`= (-220KJ)/(2moles)* 0.5021mol=-55.23 kJ`

The ΔH of the reaction when 48.7 grams of ZnS reacts with oxygen is -55.23 kJ.