posted by Mona .
Referring to the following reaction:
Pb(NO3)2 (aq) + Na2SO4 (aq) -->PbSO4 (s) + 2NaNO3 (aq)
A solution of 0.114 M Pb(NO3)2 with a volume 10.6 mL is added to 12.4 mL of a solution of 0.105 M Na2SO4. Calculate the number of grams of PbSO4 that will form.
I already checked out the equation and it is balanced, converted 10.6 mL to 0.0106 L and 12.4 mL to 0.0124 L but I just need to know if this is a simple finding mass using volume and density?
I have tried using (0.0106 L)*(0.114 M Pb(No3)2)/1 L to get 1.2084x10^-3 and (0.0124 L)*(0.105 M Na2SO4)/1 L to get 1.302x10-3. I know this is also a limiting reactant problem as well but how can I reduce the numbers because I feel like it might not be the answer my professor's looking for.
Then would it be mass of solute over mass of solution (moles over g/mol) to find the grams?
I'm just a little lost is all.
I think you are on the right track. You just have another step or two to go. And you're right, this is a limiting reagent problem; I usually work these the long way.
mols Pb(NO3)2 = M x L = approx 0.00121
mols Na2SO4 = approx 0.00130
Pb(NO3)2 x Na2SO4 ==> PbSO4 + 2NaNO3
Next, I determine the limiting reagent.
I convert 0.00121 mol Pb(NO3)2 to PbSO4 ASSUMING I have all of the Na2SO4 needed. Use the coefficients in the balanced equation to do that. That's
0.00121 x [1 mol PbSO4/1 mol Pb(NO3)] = 0.00121 mols PbSO4.
Then I convert 0.00130 mols Na2SO4 to mols PbSO4 ASSUMING I have all of the Pb(NO3)2 needed. That's
0.00130 mol Na2SO4 x (1 mol PbSO4/1 mol Na2SO4) = 0.00130 mols PbSO4.
Obviously, both answers can't be right. The correct value in limiting reagent problems is ALWAYS the smaller number and the reagent responsible for that is the limiting reagent.
Therefore, 0.00121 mols PbSO4 will be formed. Convert to grams by g = mols x molar mass = ?