Limiting reactants and excess reactants :: GCSE Chemistry Coursework Investigation

Limiting reactants and excess reactants In the first experiment we noticed how Phenolphthalein, thiosulfate and copper (II) sulfate changed their physical properties once mixed with NaOH, Iodine and Ammonia I. INTRODUCTION A chemical reaction is a change that takes place when two or more substances (reactants) interact to form new substances (products). In a chemical reaction, not all reactants are necessarily consumed. One of the reactants may be in excess and the other may be limited. The reactant that is completely consumed is called limiting reactant, whereas unreacted reactants are called excess reactants. Amounts of substances produced are called yields. The amounts calculated according to stoichiometry are called theoretical yields whereas the actual amounts are called actual yields. The actual yields are often expressed in percentage, and they are often called percent yields. In this experiment we combined sulfuric acid and aqueous barium chloride to produce a precipitate, barium sulfate and hydrochloric acid. The precipitation was isolated by filtration and theoretical yield was calculated. We predicted the limiting reactant and verified our hypothesis in the lab. II. RESULT ANALYSIS GRAPH II. DISCUSSION In this experiment we combined sulfuric acid and aquenous barium chloride to produce a precipitate, barium sulfate, and hydrochloric acid. Our assigned volumes of 0.20 M BaCl were 5mL and 30mL. H SO + BaCl BaSO + 2HCl After finishing the experiment we calculate the mass of BaSO that we isolated. The results of the two trials were: 0.7g when we used 30 mL of BaCl and 0.017g when we used 5 mL of BaCl. 1. We calculated the theoretical yield of BaSO using our assigned volume. We know that: Molarity= # of moles/ # of liters, so: Trial 1. To find the number of moles we use the molarity formula: 30mL= 0.03L 0.2M = #of moles/ 0.03L = 0.006 moles of BaCl We know from the chemical formula that there is a 1/1 mole ratio between BaCl and BaSO, and that AW of 1 mol of BaSO = 233.404, so we transform moles to grams: 0.006 x (233.404g) =1.400g BaSO Trial 2. To find the no. of moles we used the molarity formula: 5.0 mL = 0.005L 0.2M = # of moles / 0.005 = 0.001 moles of BaCl AW of 1 mole of BaSO = 233.404g, so we transform moles to grams: 0.001 x (233.404g) = 0.233g BaSO 2. After determining the theoretical yield we calculated the percent yield of BaSO: Trial 1. The actual mass of BaSO isolated in our experiment was 0.