Question

Given the following reaction:

2 ZnS(s) + 3 O2(g) → 2 ZnO(s) + 2 SO2(g) ΔH = -927.54 kJ

How much energy would be released if 50.0g of ZnS(s) is reacted with 30.0g of O2(g)?

Answer 1

Answer: -354.78 kJ of energy would be released for a given amount.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     ......(1)

For Zinc:

Mass of zinc = 50.0 g

Molar mass of zinc = 65.38 g/mol

Plugging values in equation 1:

$\text{Moles of Zn}=\frac{50.0g}{65.38g/mol}=0.765mol$

For oxygen gas:

Mass of oxygen gas = 30.0 g

Molar mass of oxygen gas = 32 g/mol

Plugging values in equation 1:

$\text{Moles of }O_2=\frac{30.0g}{32g/mol}=0.9375mol$

The chemical equation follows:

$2ZnS(s)+3O_2(g)\rightarrow 2ZnO(s)+2SO_2(g);\Delta H=-927.54kJ$

By the stoichiometry of the reaction:

If 2 moles of zinc reacts with 3 moles of oxygen gas

So, 0.765 moles of zinc will react with = $\frac{3}{2}\times 0.765=0.51mol$ of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, Zn is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of zinc are releasing 927.54 kJ of energy

So, 0.765 moles of zinc will react with = $\frac{927.54}{2}\times 0.765=354.78kJ$ of energy

Hence, -354.78 kJ of energy would be released for a given amount.