Question

Consider the following reaction:
3Pb(NO3)2(aq) +Al2(SO4)3(aq) ¡÷ 3PbSO4(aq) + 2Al(NO3)3(aq)

Determine the mass (in gram) of lead(II) sulphate (PbSO4) and aluminium nitrate [Al(NO3)3] formed in the solution when 8.84 g of lead(II) nitrate [Pb(NO3)2] was reacted with 6.28 g of aluminium shlphate [Al2(SO4)3].

Answer 1

This is a limiting reagent (LR) problem and you know that because amounts for BOTH reactants is given.
mols Pb(NO3)2 = grams/molar mass = ?
mols Al2(SO4)3 = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols Pb(NO3)2 to mols either product.
Do the same and convert mols Al2(SO4)3 to mols of the SAME product.
It is likely these values will not agree which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that value is the LR. It is essential at this step to identify the LR. Using the smaller value of the product, convert to grams. g = mols x mlar mass.

For the other product, use the LR and convert with the coefficients to determine mols of the other product and convert to grams as above.