Limiting Reactants and Percent Yield. AP Chemistry. What is a Limiting Reactant?. It is the reactant in a reaction that determines how much product can be made. It is whatever reactant you have the least amount of.
1. Write a balanced equation.
2. Do a separate mass to mass problem starting with each reactant. The smaller answer is correct.
To find out how much of the excess reactant is left over,
1. Start with the initial mass of the limiting reactant and
2. Do a mass to mass problem to determine how much of the excess reactant was needed.
3. Subtract that value from the initial mass of the excess reactant.
1. What volume of hydrogen gas at STP is produced from the reaction of 50.0g of Mg and 75.0 grams of HCl? How much of the excess reagent is left over (in grams)?
Mg(s) + HCl(aq) MgCl2(s) + H2(g)
Do a standard mass to mass problem starting with each reactant
50.0 gMg 1mol Mg 1mol H2 22.4 L
24.31g Mg 1mol Mg 1mol H2
= 46.1 L H2
75.0 g HCl 1mol HCl 1 mole H2 22.4 L
36.46 g HCl 2mol HCl 1mol H2
= 23.0 L H2
HCl is the limiting reactant!!
36.46g HCl 2mol HCl 1mol Mg
= 25.0 grams Mg needed for the reaction
Thus, 50.0 grams – 25.0 grams =
25.0 grams Mg leftover!
2. What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid (98% H2SO4 by mass)?
Ca3(PO4) 2 (s) + H2SO4(aq) CaSO4(s) + H3PO4(aq)
1000g Ca3(PO4)2 1mol Ca3(PO4)2 3mol CaSO4 136.15g CaSO4
310.18g 1 mol Ca3(PO4)2 1mol CaSO4
= 1317 g
980 g H2SO4 1mol H2SO4 3mol CaSO4 136.15g CaSO4
98.09g H2SO4 3mol H2SO4 1mol CaSO4
= 1360 g
Write balanced reaction Cu + S Cu2S
Determine theoretical yield – doing a mass to mass problem
1.50g Cu 1 mol Cu 1mol Cu2S 159.17g Cu2S
63.55g Cu 2 mol Cu 1mol Cu2S
= 1.88 g Cu2S
Percent Yield = 1.76 g x 100 = 93.6 %