Question

A student reacted 10.0 grams of NaOH with 12.0 grams of HCl to produce NaCl and H2O. The balanced chemical equation for the reaction is as follows:NaOH + HCl → NaCl + H2OIf the expected yield of NaCl is 9.0 grams, what is the percent yield of the reaction?

Answer 1

First, we need to determine the limiting reactant. To do this, we will compare the moles of NaOH and HCl used in the reaction.Molar mass of NaOH = 22.99  Na  + 15.999  O  + 1.00784  H  = 39.996 g/molMoles of NaOH = 10.0 g / 39.996 g/mol = 0.250 molMolar mass of HCl = 1.00784  H  + 35.453  Cl  = 36.461 g/molMoles of HCl = 12.0 g / 36.461 g/mol = 0.329 molSince the balanced equation has a 1:1 ratio between NaOH and HCl, NaOH is the limiting reactant because it has fewer moles  0.250 mol  compared to HCl  0.329 mol .Now, we can calculate the theoretical yield of NaCl based on the limiting reactant  NaOH .Molar mass of NaCl = 22.99  Na  + 35.453  Cl  = 58.443 g/molTheoretical yield of NaCl = 0.250 mol * 58.443 g/mol = 14.611 gNow, we can calculate the percent yield using the expected yield  9.0 g  and the theoretical yield  14.611 g .Percent yield =  Expected yield / Theoretical yield  * 100Percent yield =  9.0 g / 14.611 g  * 100 = 61.6%The percent yield of the reaction is 61.6%.