Question

Using the balanced equation below,

Answer 1

Answer: The mass of manganese(III) oxide produced is 113.03 g

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)

Given mass of zinc = 46.8 g

Molar mass of zinc = 65.38 g/mol

Plugging values in equation 1:

$\text{Moles of zinc}=\frac{46.8g}{65.38g/mol}=0.716 mol$

The given chemical equation follows:

$Zn+2MnO_2+H_2O\rightarrow Zn(OH)_2+Mn_2O_3$

By the stoichiometry of the reaction:

If 1 mole of zinc produces 1 mole of manganese(III) oxide

So, 0.716 moles of zinc will produce = $\frac{1}{1}\times 0.716=0.716mol$ of manganese(III) oxide

Molar mass of manganese(III) oxide = 157.87 g/mol

Plugging values in equation 1:

$\text{Mass of manganese(III) oxide}=(0.716mol\times 157.87g/mol)=113.03g$

Hence, the mass of manganese(III) oxide produced is 113.03 g