Chemistry Worksheet: Limiting Reactant Worksheet #1 1. Consider the following reaction: 2 Al + 6 HBr → 2 AlBr3 + 3 H2 a. When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H2 are formed? b. What is the limiting reactant? c. For the reactant in excess, how many moles are left over at the end of the reaction? 2. Consider the following reaction: 3 Si + 2 N2 → Si3N4 a. When 21.44 moles of Si reacts with 17.62 moles of N2, how many moles of Si3N4 are formed? b. What is the limiting reactant? c. For the reactant in excess, how many moles are left over at the end of the reaction? 3. Consider the following reaction: 2 CuCl2 + 4 KI → 2 CuI + 4 KCl + I2 a. When 0.56 moles of CuCl2 reacts with 0.64 moles of KI, how many moles of I2 are formed? b. What is the limiting reactant? c. For the reactant in excess, how many moles are left over at the end of the reaction? 4. Consider the following reaction: 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 a. When 26.62 moles of FeS2 reacts with 5.44 moles of O2, how many moles of SO2 are formed? b. What is the limiting reactant? c. For the reactant in excess, how many moles are left over at the end of the reaction?Chemistry Worksheet: Limiting Reactant Worksheet #1 1. Consider the following reaction: 2 Al + 6 HBr → 2 AlBr3 + 3 H2 a. When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H2 are formed? 2.48 mol H2 b. What is the limiting reactant? HBr c. For the reactant in excess, how many moles are left over at the end of the reaction? 1.57 mol Al 3.22 mol Al * (3 mol H2 / 2 mol Al) = 4.83 mol H2 4.96 mol HBr * (3 mol H2 / 6 mol HBr) = 2.48 mol H2 3.22 mol Al 2.48 mol H2 * (2 mol Al / 3 mol H2) = 1.65 mol Al used up –1.65 mol Al 1.57 mol Al 2. Consider the following reaction: 3 Si + 2 N2 → Si3N4 a. When 21.44 moles of Si reacts with 17.62 moles of N2, how many moles of Si3N4 are formed? 7.147 mol Si3N4 b. What is the limiting reactant? Si c. For the reactant in excess, how many moles are left over at the end of the reaction? 3.33 mol N2 21.44 mol Si * (1 mol Si3N4 / 3 mol Si) = 7.147 mol Si3N4 17.62 mol N2 * (1 mol Si3N4 / 2 mol N2) = 8.810 mol Si3N4 21.44 mol N2 7.147 mol Si3N4 * (2 mol N2 / 1 mol Si3N4) = 14.29 mol N2 –14.29 mol N2 used up 3.33 mol N2 3. Consider the following reaction: 2 CuCl2 + 4 KI → 2 CuI + 4 KCl + I2 a. When 0.56 moles of CuCl2 reacts with 0.64 moles of KI, how many moles of I2 are formed? 0.16 mol I2 b. What is the limiting reactant? KI c. For the reactant in excess, how many moles are left over at the end of the reaction? 0.24 mol CuCl2 0.56 mol CuCl2 * (1 mol I2 / 2 mol CuCl2) = 0.28 mol I2 0.64 mol KI * (1 mol I2/ 4 mol KI) = 0.16 mol I2 0.56 mol CuCl2 0.16 mol I2 * (2 mol CuCl2 / 1 mol I2) = 0.32 mol CuCl2 –0.32 mol CuCl2 used up 0.24 mol CuCl2 4. Consider the following reaction: 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 a. When 26.62 moles of FeS2 reacts with 5.44 moles of O2, how many moles of SO2 are formed? 3.96 mol SO2 b. What is the limiting reactant? O2 c. For the reactant in excess, how many moles are left over at the end of the reaction? 24.64 mol FeS2 26.62 mol FeS2 * (8 mol SO2 / 4 mol FeS2) = 53.24 mol SO2 5.44 mol O2 * (8 mol SO2/ 11 mol O2) = 3.96 mol SO2 26.62 mol FeS2 3.96 mol SO2 * (4 mol FeS2 / 8 mol SO2 )= 1.98 mol FeS2 –1.98 mol FeS2 used up 24.64 mol
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