Question

consider the reaction: zno(s) + h2o(l) → zn(oh)2(aq) given an initial mass of 16.08 g zno, an excess of h2o, and assuming that all of the reactant is converted to product(s), and none is lost, calculate the mass (g) of zn(oh)2 produced by the reaction.

Explanation:

Step1: Calculate molar masses

The molar mass of $ZnO$ is $M_{ZnO}=65.38 + 16.00=81.38\ g/mol$. The molar mass of $Zn(OH)_2$ is $M_{Zn(OH)_2}=65.38+2\times(16.00 + 1.01)=99.40\ g/mol$.

Step2: Determine moles of $ZnO$

The number of moles of $ZnO$, $n_{ZnO}=\frac{m_{ZnO}}{M_{ZnO}}=\frac{16.08\ g}{81.38\ g/mol}=0.1976\ mol$.

Step3: Use mole - ratio from the balanced equation

The balanced chemical equation is $ZnO(s)+H_2O(l)
ightarrow Zn(OH)_2(aq)$. The mole - ratio of $ZnO$ to $Zn(OH)_2$ is 1:1. So, the number of moles of $Zn(OH)_2$ produced, $n_{Zn(OH)_2}=n_{ZnO}=0.1976\ mol$.

Step4: Calculate mass of $Zn(OH)_2$

The mass of $Zn(OH)_2$ produced, $m_{Zn(OH)_2}=n_{Zn(OH)_2}\times M_{Zn(OH)_2}=0.1976\ mol\times99.40\ g/mol = 19.64\ g$.

Answer:

19.64 g