A 16.0 mL sample of a 1.04 M potassium sulfate solution is mixed with 14.3 mL of a 0.880 M barium nitrate solution and this prec
ipitation reaction occurs: K 2 S O 4 (aq)+Ba(N O 3 ) 2 (aq)→BaS O 4 (s)+2KN O 3 (aq) The solid BaS O 4 is collected, dried, and found to have a mass of 2.60 g . Determine the limiting reactant, the theoretical yield, and the percent yield.
Number of moles in the K2SO4 sample = (16/1000)*1.04= 0.01664 mol
Number of moles in the Ba(NO3)2 sample = (14.3/1000*0.880)= 0.01258 mol
Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.
The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4 Therefore the theoretical yield of Barium Sulphate is 233.4*0.01258=2.937 g Actual yield = 2.60 g (given) Therefore the percentage yield = 2.60/2.937=88.54%
Answer: 1. the limiting reagent is Barium Nitrate (Ba(NO3)2) 2. the theoretical yield is 2.94 g 3. the percentage yield is 88.5%
I apologize for the mistake previous to this update.
