If 0.10 mol of Zn(s) is reacted to 150.0 ml of 1M HCl, what is the concentration of Zn+2 in the solution after the reaction is complete? Assume the solid zinc does not affect the volume.
Zn + 2HCl ==> ZnCl2 + H2
mols Zn = 0.10
mols HCl = M x L = 1 x 0.150 = 0.150
First you must determine the limiting reagent (LR). You know this is a LR problem because amounts are given for BOTH reactants.
a. Convert 0.10 mol Zn to mols H2. That's 0.10 mol Zn x (1 mol H2/1 mol Zn) = 0.10 mol H2.
b. Do the same for HCl. That's 0.150 mols HCl x (1 mol H2/2 mol HCl) = 0.075 mols H2.
c. Note that the two values don't agree which means one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR. So HCl is the limiting reagent and all of it will react.
How much Zn will it use? That's 0.150 mol HCl x (1 mol Zn/2 mol HCl) = 0.075 mol Zn used. We had 0.1 initially; we used 0.075 mol so that's how much Zn^2+ is formed. What's the volume? That's 150 mL; therefore, (Zn^2+) = mols/L = ?
And there is no reason to assume the solid Zn will not affect the final volume because it won't.