Limiting reactants and excess reactants :: GCSE Chemistry Coursework Investigation

Limiting reactants and excess reactantsIn the first examine we noticed how Phenolphthalein, thiosulfate and copper (II) sulfate changed their physical properties once blend with NaOH, Iodine and AmmoniaI. INTRODUCTIONA chemical reaction is a change that takes place when two or moresubstances (reactants) interact to form new-fashioned 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 amountscalculated according to stoichiometry are called theoretical yieldswhereas the actual amounts are called actual yields. The actual yieldsare a great deal expressed in percentage, and they are often called percentyields.In this experiment we combined sulfuric acid and aqueous atomic number 56chloride to produce a p recipitate, barium sulfate and hydrochloricacid. The precipitation was isolated by filtration and theoreticalyield was calculated. We predicted the limiting reactant and verifiedour hypothesis in the lab.II. RESULT 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 + 2HClAfter finishing the experiment we calculate the people 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 tell apart thatMolarity= of moles/ of liters, soTrial 1.To find the number of moles we use the molarity formula30mL= 0.03L0.2M = of moles/ 0.03L = 0.006 moles of BaClWe know from the chemical formula that there is a 1/1 mole ratiobetween BaCl and BaSO, and that AW of 1 mol of BaSO = 233.404, so wetransform 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.