Limiting reagents • Definition: limits the amount of product that can be formed in a reaction • The reaction occurs until the limiting reagent is used up • Excess reagent = not completely used up
N2 + 3H2 2NH3 LIMITING REAGENT
Suppose that 6.70 mol Na reacts with 3.20 mol Cl2. 2 Na(s) + Cl2 (g) 2 NaCl (s) • What is the limiting reagent? • How many moles of NaCl are produced? Choose one of the reagents and calculate how much of the other reagent will be needed. 6.70 mol Na 3.35 mol of Cl2 is needed, which is more than the 3.20 mol provided in the problem. Therefore, Cl2 is the limiting reagent because we would run out of it. 1 mol Cl = 3.35 mol Cl2 2 mol Na
2 Cu(s) + S (s) Cu2 S (s) • What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S? • What is the maximum number of grams of Cu2 S? Convert grams to moles for given quantities. 80.0g Cu 25.0 g S LIMITING REAGENT 1 mol Cu = 1.26 mol Cu 63.5 g Cu 1 mol S = 0.779 mol S 32.1 g S Requires less Sulfur 1 mol S 1.26 mol Cu = 0.630 mol S 2 mol Cu
2 Cu(s) + S (s) Cu2 S (s) • What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S? • What is the maximum number of grams of Cu2 S? Convert moles to grams of product using mole ratio. 1.26 mol Cu 1 mol Cu2S 159.1gCu2S 1 mol Cu2S 2 mol Cu LIMITING REAGENT = 100 g Cu2S
Calculating percent yield • Theoretical yield = max. amt of product that could be obtained from given amt of reactants • Actual yield = the amt of product that actually is formed when the reaction is carried out in a lab • Percent yield = ratio of actual yield to theoretical yield
…the equation actual yield % yield = theoretical yield X 100
CaCO3 CaO + CO2 • What is the theoretical yield of CaO if 24.8g is heated? • What is the percent yield if 13.1 g CaO is produced?