Answer:
[tex]m_{Zn(OH)_2}=38.4g[/tex]
Explanation:
Hello!
In this case, for the undergoing chemical reaction:
[tex]ZnO(s)+H_2O(l)\rightarrow Zn(OH)_2[/tex]
We evaluate the yielded moles of zinc hydroxide by each reactant as shown below:
[tex]n_{Zn(OH)_2}^{by ZnO}=35.4gZnO*\frac{1molZnO}{81.38gZnO}*\frac{1molZn(OH)_2}{1molZnO} =0.435molZn(OH)_2\\\\n_{Zn(OH)_2}^{by H_2O}=6.96gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{1molZn(OH)_2}{1molH_2O} =0.386molZn(OH)_2[/tex]
In such a way, since the water yields a smaller amount of zinc hydroxide we conclude it is the limiting reactant so the maximum mass is computed below:
[tex]m_{Zn(OH)_2}=0.386molZn(OH)_2*\frac{99.424 gZn(OH)_2}{1molZn(OH)_2} \\\\m_{Zn(OH)_2}=38.4g[/tex]
Because the water limits the yielded amount of zinc hydroxide.
Best regards!
