If 0.188 moles of zinc reacts with excess lead(IV) sulfate, how many grams of zinc sulfate would be produced in the following reaction?
To solve this problem, you need to use the concept of stoichiometry and the balanced equation for the reaction.
The balanced equation for the reaction between zinc and lead(IV) sulfate is:
Zn + Pb(SO4)2 → ZnSO4 + Pb
According to the reaction equation, the stoichiometric ratio between Zn and ZnSO4 is 1:1. This means that for every 1 mole of zinc (Zn), you will produce 1 mole of zinc sulfate (ZnSO4).
Given that you have 0.188 moles of zinc, you can conclude that you will produce 0.188 moles of zinc sulfate.
To find the mass of zinc sulfate, you need to use the molar mass of ZnSO4, which can be calculated as follows:
Molar mass of Zn = 65.38 g/mol
Molar mass of S = 32.07 g/mol
Molar mass of O (4 oxygen atoms) = 16.00 g/mol x 4 = 64.00 g/mol
Molar mass of ZnSO4 = 65.38 g/mol + 32.07 g/mol + 64.00 g/mol = 161.43 g/mol
Now, you can use the molar mass of zinc sulfate and the number of moles calculated earlier to find the mass of zinc sulfate:
Mass of ZnSO4 = Number of moles of ZnSO4 x Molar mass of ZnSO4
Mass of ZnSO4 = 0.188 mol x 161.43 g/mol = 30.34 g
Therefore, if 0.188 moles of zinc reacts with excess lead(IV) sulfate, it would produce 30.34 grams of zinc sulfate.