Using the balanced equation above, determine the amount of excess reactant remaining when 48.1 g of Na2SO4 reacts with 75.0 g of AgNO3.
1. Write down the balanced chemical equation:
2AgNO3 + Na2SO4 -> Ag2SO4 + 2NaNO3
2. Convert the masses of the reactants into moles:
48.1 g Na2SO4 x (1 mol Na2SO4 / 142.04 g Na2SO4) = 0.339 mol Na2SO4
75.0 g AgNO3 x (1 mol AgNO3 / 169.87 g AgNO3) = 0.442 mol AgNO3
3. Determine the limiting reactant by comparing the moles of each reactant to the stoichiometry of the balanced equation. In this case, AgNO3 is the limiting reactant because it requires 2 moles of AgNO3 to react with 1 mole of Na2SO4.
4. Determine the theoretical yield of the product formed:
0.442 mol AgNO3 x (1 mol Ag2SO4 / 2 mol AgNO3) = 0.221 mol Ag2SO4
5. Calculate the moles of excess reactant remaining after the reaction:
0.339 mol Na2SO4 - (0.221 mol Ag2SO4 / 1 mol Ag2SO4) = 0.118 mol Na2SO4
6. Convert the moles of excess reactant into grams:
0.118 mol Na2SO4 x (142.04 g Na2SO4 / 1 mol Na2SO4) = 16.77 g Na2SO4
Therefore, 16.77 g of Na2SO4 will be remaining after the reaction is complete.