Limiting Reactants It will be necessary to determine the limiting reactant whenever we are given the amounts of two or more reactants in a chemical reaction. The limiting reactant limits the amount of product that can be obtained and is therefore totally consumed in the reaction. Any other reactant is in excess. Problem I: mole Given: 9.25 moles of potassium chloride and 8.75 moles of chromium (III) chloride react according to the following equation: mole 2 KCl(aq) + 2 CrCl3(aq) 7H2O(l) + 3 Cl2(g) Problem: 1. find the limiting reactant 2. Find the amount of chlorine formed 3. Find the amount of excess Solution: 1. Compare moles of each reactant to moles of product (Cl2) 2. The amount of chlorine formed is 13.1 mol. 3. Amount of excess used (Look back at the original moles of reactants.) n = 8.75 moles KCl(aq) used up Excess = original moles KCl(aq) - moles used KCl(aq) Excess = 9.25 moles - 8.75 moles Excess = 0.50 moles Problem II: Volume/concentration moles mass Given: 348 mL of 1.50 Ba(OH)2(aq) is added to 285 mL of 2.30 and reacts according to the following equation: 3Ba(OH)2(aq) + 2H3P04(aq) Ba3(P04)2(aq) + 6H2O(l) Problem: 1. Find the limiting reactant. 2. Find the mass of barium phosphate formed. 3. Find the mass of reagent in excess. Solution: 1. 3Ba(OH)2(aq) + 2H3P04(aq) of H3PO4(aq) Ba3(P04)2(aq) + 6H2O(l) V = 348 mL = 0.348 L V = 285 mL = 0.285 L C = 1.50 n=CV C = 2.30 n=CV n = (1.50 )(0.348 L ) n = 0.522 mol n = (2.30 )(0.285 L) n = 0.6555 mol n = 0.32775 mol This is the smaller amount, therefore, Ba(OH)2(aq) is the limiting reactant 2. The mass of Ba3(P04)2(aq) formed is 0.1 74 mol x 3. Excess used up n = 0.116 mol Excess = original moles - moles used up of H3P04(aq) Excess = 0.6555 mol - 0.116 mol Excess = 0.5395 mol Excess = 0.5395 mol x Method gram I mole gram Given: 4.20 grams of sulphuric acid and 7.30 grams of sodium hydroxide react according to the equation: H2SO4(aq) +2NaOH(aq) Problem: Na2 S04(aq) + 2H2O(l) 1. Find the limiting reactant. 2. Find the amount of product formed. Na2 S04(aq) 3. Find the amount of excess. Solution: H2SO4(aq) +2NaOH(aq) Na2 S04(aq) + 2H2O(l) 1. a. b. compare mole of reactants to moles of product This is the smaller amount, therefore, H2SO4(aq) is the limiting reactant. 2. a. The amount of product formed is 4.28 x 10-2 moles Na2 S04(aq) b. Amount of product in grams is 4.28 x 10-2 moles x 142.04/mol = 6.08 g 3. a. Amount of excess used (go back to original moles and look at reactants) x = 8.56 x 10-2 moles NaOH(aq) used b. Excess = original moles NaOH(aq) - molesNaOH(aq) used Excess = 1.83 x 10-1 moles - 8.56 x 10-2 moles Excess = 9.74 x 10-2 moles NaOH(aq) c. Amount of excess in grams 9.74 x 10-2 moles x NaOH(aq) = 3.90 g Method II gram Given: 4.20 grams of sulphuric acid and 7.30 grams of sodium hydroxide react according to the equation: gram H2SO4(aq) +2NaOH(aq) Problem: Na2 S04(aq) + 2H2O(l) 1. Find the limiting reactant. 2. Find the amount of product formed. Na2 S04(aq) 3. Find the amount of excess. Solution: compare each of the reactants to the product 1. This is the smaller amount, therefore, H2SO4(aq) is the limiting reactant. 2. The amount of product formed is 6.08 g. 3. a. Amount of excess (go back to original masses and look at the reactants). x = 3.43 g (NaOH used up) b. Excess = original mass - mass used up Excess = 7.30 g - 3.43 g Excess = 3.90 g Using the following equation, calculate the concentration of each after the solutions are mixed. a. b. c. [Ca2+(aq)] [NO3-(aq)] [Ag+(aq)] CaCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Ca(N03)2(aq) V = 25.0 mL V = 20.0 mL C = 1.78 C = 6.50 n = 0.0445 mol n = 0.130 mol Ca2+(aq) + 2NO3-(aq) Ca(NO3)2(aq) [Ag+(aq)] Since AgNO3(aq) is in excess part of the Ag will go into solution as ions, the rest will form the solid AgCl(s). Find the amount used up and the excess. n = 0.0890 mol AgNO3(aq) used up Excess = 0.130 mol - 0.089 mol = 0.0410 mol Exercises A. CaCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Ca(N03)2(aq) Given: 4.32 moles of calcium chloride react with 8.75 moles of silver nitrate. Calculate the amount of silver chloride produced. B. 2C4H10(g) + 1302(g) 8CO2(g) + 10H2 O(g) Given: 20.0 L of butane reacts with 90.0 L of oxygen gas at STP. Find the volume of carbon dioxide gas formed. C. 2NaI(aq) + Pb(NO3)2(aq) PbI2(s) + 2NaNO3(aq) sodium iodide reacts with 4.58 L of 0.500 Given: 3.79 L of 1.00 nitrate. Calculate the mass of precipitate produced. For each of the following find: 1. the limiting reactant 2. the amount of product formed in moles and grams 3. the amount of excess in moles and grams D. 15.0 g of iron (III) oxide reacts with 7.00 g of carbon monoxide. Find the amount of Iron formed. Fe2O3(s) + 3CO(g) 2 Fe(s) + 3CO2(g) E. 7.5 g of sodium chloride reacts with 9.8 g of Lead (II) nitrate. Find the amount of Lead Chloride formed. 2NaCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2NaNO3(aq) F. 57 g of magnesium hydroxide is added to 86 g of hydrogen phosphate. Find the amount of Magnesium phosphate formed. 3Mg(OH)2(aq) + 2H3PO4(aq) Mg3(P04)2(s) + 6H2 O(l) lead (II)