Question

Using the balanced equation below,

how many grams of manganese(III)
oxide would be produced from the
complete reaction of 46.8 g of zinc?
Zn + 2MnO2 + H20 — Zn(OH)2 + Mn203

Answer 1

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

Step-by-step 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}=(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 =
`(1)/(1)* 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* 157.87g/mol)=113.03g`

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