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:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of zinc = 46.8 g
Molar mass of zinc = 65.38 g/mol
Plugging values in equation 1:
[tex]\text{Moles of zinc}=\frac{46.8g}{65.38g/mol}=0.716 mol[/tex]
The given chemical equation follows:
[tex]Zn+2MnO_2+H_2O\rightarrow Zn(OH)_2+Mn_2O_3[/tex]
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 = [tex]\frac{1}{1}\times 0.716=0.716mol[/tex] of manganese(III) oxide
Molar mass of manganese(III) oxide = 157.87 g/mol
Plugging values in equation 1:
[tex]\text{Mass of manganese(III) oxide}=(0.716mol\times 157.87g/mol)=113.03g[/tex]
Hence, the mass of manganese(III) oxide produced is 113.03 g
