here u goExplanation:Use the following reaction: C4H9OH + NaBr + H2SO4 C4H9BR + NaHSO4 + H2O If 15.0 G of C4H9OH react with 22.4 g of NaBr and 32.7 g of H2SO4 to yield 17.1 g of C4H9Br, what is the percent yield of this reaction? Remember: percent yield is your (experimental yield/theoretical yield)x100. Answer: The percent yield of the reaction is 61.5 %.Explanation:To calculate the number of MOLES, we use the equation:      .....(1)For butanol:GIVEN mass of butanol = 15.0 gMolar mass of butanol = 74 g/molPutting values in equation 1, we get:For NaBr:Given mass of NaBr = 22.4 gMolar mass of NaBr = 103 g/molPutting values in equation 1, we get:For sulfuric ACID:Given mass of sulfuric acid = 32.7 gMolar mass of sulfuric acid = 98 g/molPutting values in equation 1, we get:For the given chemical reaction:As, the reactants are present in 1 : 1 : 1 ratio. So, the reactant having minium number of moles will be considered as the limiting reagent.Here, the limiting reagent is butanol because it limits the formation of product.By Stoichiometry of the reaction:1 mole of butanol produces 1 mole of bromobutaneSo, 0.203 moles of butanol will produce =  of bromobutaneNow, calculating the mass of bromobutane from equation 1, we get:Molar mass of bromobutane = 137 g/molMoles of bromobutane = 0.203 molesPutting values in equation 1, we get:To calculate the percentage yield of bromobutane, we use the equation:Experimental yield of bromobutane = 17.1 gTheoretical yield of bromobutane = 27.81 gPutting values in above equation, we get:Hence, the percent yield of the reaction is 61.5 %.