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

Answer:

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

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.:

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

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

So, 0.5021 moles of ZnS gives energy :

$= \frac{-220KJ}{2moles}\times 0.5021mol=-55.23 kJ$

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

Answer 2

Answer : The $\Delta H_{rxn}$ for the reaction is, 54.89 KJ

Solution : Given,

Mass of ZnS = 48.7 g

Molar mass of ZnS = 97.474 g/mole

$\Delta H=220KJ$

First we have to calculate the moles of ZnS.

$\text{ Moles of ZnS}=\frac{\text{ Mass of ZnS}}{\text{ Molar mass of ZnS}}=\frac{48.7g}{97.474g/mole}=0.499moles$

The balanced combustion reaction is,

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

From the given reaction, we conclude that

As, 2 moles of ZnS gives energy = 220 KJ

So, 0.499 moles of ZnS gives energy = $\frac{220KJ}{2moles}\times 0.499moles=54.89KJ$

Therefore, the $\Delta H_{rxn}$ for the reaction is, 54.89 KJ