Question

aqeous HRb reacts with NaOH to produce aqeous NaBr and liquid water. If 1.12 g of sodium bromide is produced from the reaction of 1.6 g of hydrobromic acid and 1.2 grams of sodium hydroxide calculate the percent yield of sodium bromide. 2 sig figs

Answer 1

Mass of NaBr = 1.12 g Mass of HBr = 1.6 g Mass of NaOH = 1.2 g Moles of HBr = 1.6/80.91 = 0.01977 moles Moles of NaOH = 1.2/39.997 = 0.0300 moles Since moles of NaOH > moles of HBr Therefore HBr is limiting reagent Theoretical yield = moles of limiting * molar mass of NaBr = 0.01977 * 102.894 = 2.04 g % yield = actual yield/Theoretical yield * 100 = 1.12/2.04 * 100 = 54.90% % yield = 54.90% almostequalto 55% (2 sig. figures)