Question

How many grams of zinc would be required to produce 9.65g of zinc hydroxide

Answer 1

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:

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

$\text{Moles of zinc hydroxide}=\frac{9.65g}{99.4g/mol}=0.0971mol$

The given chemical equation follows:

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

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 = $\frac{1}{1}\times 0.0971=0.0971mol$ 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:

$0.0971mol=\frac{\text{Mass of zinc}}{65.4g/mol}\\\\\text{Mass of zinc}=(0.0971mol\times 65.4g/mol)=6.35g$

Hence, the mass of zinc required is 6.35 grams