Limiting reactants and excess reactants :: GCSE Chemistry Coursework Investigation

Limiting reactants and excess reactantsIn the first audition we noticed how Phenolphthalein, thiosulfate and copper (II) sulfate swopd their physical properties once mixed with NaOH, one and AmmoniaI. INTRODUCTIONA chemical reaction is a change that takes place when two or moresubstances (reactants) interact to form clean substances (products).In a chemical reaction, not all reactants are of necessity 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 metrical according to stoichiometry are called theoretical yieldswhereas the effective amounts are called actual yields. The actual yieldsare often expressed in portionage, and they are often called percentyields.In this experiment we combined sulphuric acid and aqueous bariumchloride to produce a precipitate, barium sul fate and hydrochloricacid. The precipitation was isolated by filtration and theoreticalyield was calculated. We predicted the limiting reactant and verifiedour hypothesis in the lab.II. direct ANALYSIS GRAPHII. DISCUSSIONIn this experiment we combined sulfuric acid and aquenous bariumchloride to produce a precipitate, barium sulfate, and hydrochloricacid. Our assigned volumes of 0.20 M BaCl were 5mL and 30mL.H SO + BaCl BaSO + 2HCl subsequently finishing the experiment we calculate the mickle of BaSO that weisolated. The results of the two trials were0.7g when we used 30 mL of BaCl and0.017g when we used 5 mL of BaCl.1. We calculated the theoretical yield of BaSO using our assigned volume.We know that chiliad= of moles/ of liters, so rill 1.To find the number of moles we use the molarity conventionality30mL= 0.03L0.2M = of moles/ 0.03L = 0.006 moles of BaClWe know from the chemical linguistic rule that there is a 1/1 mole ratiobetween BaCl and B aSO, and that AW of 1 mol of BaSO = 233.404, so we render moles to grams0.006 x (233.404g) =1.400g BaSOTrial 2.To find the no. of moles we used the molarity formula5.0 mL = 0.005L0.2M = of moles / 0.005 = 0.001 moles of BaClAW of 1 mole of BaSO = 233.404g, so we transform moles to grams0.001 x (233.404g) = 0.233g BaSO2. After determining the theoretical yield we calculated the percent yield of BaSOTrial 1.The actual mass of BaSO isolated in our experiment was 0.