Question

For the following reaction, 35.4 grams of zinc oxide are allowed to react with 6.96 grams of water . zinc oxide(s) + water(l) ------- zinc hydroxide(aq) What is the maximum mass of zinc hydroxide that can be formed?

Answer 1

`m_(Zn(OH)_2)=38.4g`

Step-by-step explanation:

Hello!

In this case, for the undergoing chemical reaction:

`ZnO(s)+H_2O(l)\rightarrow Zn(OH)_2`

We evaluate the yielded moles of zinc hydroxide by each reactant as shown below:

`n_(Zn(OH)_2)^(by ZnO)=35.4gZnO*(1molZnO)/(81.38gZnO)*(1molZn(OH)_2)/(1molZnO) =0.435molZn(OH)_2\\\\n_(Zn(OH)_2)^(by H_2O)=6.96gH_2O*(1molH_2O)/(18.02gH_2O)*(1molZn(OH)_2)/(1molH_2O) =0.386molZn(OH)_2`

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:

`m_(Zn(OH)_2)=0.386molZn(OH)_2*(99.424 gZn(OH)_2)/(1molZn(OH)_2) \\\\m_(Zn(OH)_2)=38.4g`

Because the water limits the yielded amount of zinc hydroxide.

Best regards!