Joshua H. answered • 11/07/19
BS in Chemistry with 5+ Years Laboratory Experience
This question is fundamentally about stoichiometry, so the first step is to calculate the molar masses of each compound related to the answer. Since we're calculating the required masses of Zn(OH)2, ZnSO4, and H2O, these are the compounds we need to calculate.
Zn(OH)2: (65.38 g/mol) + 2*(15.999 g/mol) + 2*(1.008 g/mol) = 97.63 g/mol
ZnSO4: (65.38 g/mol) + (32.06 g/mol) + 4*(15.999 g/mol) = 161.44 g/mol
H2O: (15.999 g/mol) + 2*(1.008 g/mol) = 18.016 g/mol
Now that we know the molar mass for each relevant compound, we can calculate the number of moles of ZnSO4 we have:
1.234 g / (161.44 g/mol) = 0.007644 mol
This is the number of moles of ZnSO4 required for the problem. From here, we use the ratios of each compound in the equation to calculate everything.
Since there is one molecule of Zn(OH)2 per molecule of ZnSO4 in this equation, we know that the same number of moles of compound is present for each. In order to calculate the mass of Zn(OH)2 required for this reaction, we multiply the molar mass by the number of moles of ZnSO4:
0.007644 mol * 97.63 g/mol = 0.7463 g Zn(OH)2
Since there are twice as many moles of H2O as ZnSO4 in this equation, we can calculate that there will be 2*0.007644 mol = 0.01529 mol of H2O. The mass of this many moles of water is:
(18.016 g/mol) * 0.01529 mol = 0.2754 g H2O.
The reason why the problem includes the bit about HSO4 in excess is because the problem depends on the reaction continuing to completion. If HSO4 was the limiting factor in this reaction, none of the calculations would be the same.
