The question is incomplete, here is the complete question:
How many grams of zinc would be required to produce 9.65g of zinc hydroxide
Zn+2MnO₂+H₂O→Zn(OH)₂+Mn₂O₃
Answer: The mass of zinc required is 6.35 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of zinc hydroxide = 9.65 g
Molar mass of zinc hydroxide = 99.4 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of zinc hydroxide}=\frac{9.65g}{99.4g/mol}=0.0971mol[/tex]
The given chemical equation follows:
[tex]Zn+2MnO_2+H_2O\rightarrow Zn(OH)_2+Mn_2O_3[/tex]
By Stoichiometry of the reaction:
1 mole of zinc hydroxide is produced from 1 mole of zinc
So, 0.0971 moles of zinc hydroxide will be produced from = [tex]\frac{1}{1}\times 0.0971=0.0971mol[/tex] of zinc
Now, calculating the mass of zinc from equation 1, we get:
Molar mass of zinc = 65.4 g/mol
Moles of zinc = 0.0971 moles
Putting values in equation 1, we get:
[tex]0.0971mol=\frac{\text{Mass of zinc}}{65.4g/mol}\\\\\text{Mass of zinc}=(0.0971mol\times 65.4g/mol)=6.35g[/tex]
Hence, the mass of zinc required is 6.35 grams
