Question

Consider the following balanced chemical equation: 2NH3(g)+ 5F2(g)- 2SO2(g)+2H20(g) Determine the theoretical yield of water, in grams, when 80.1 g of NH3 reacts with 80.1 g of F2. g H20 What is the percent yield if 9.73 g of water is actually formed during the reaction?

Answer 1

2 NH_3 (g) + 5 F_2 (g) rightarrow 2 SO_2 (g) rightarrow 2H_2 O_(g) 80.1 g 80.1 g Initial mole of NH_3 = 80.1 g/17.03 g/mol = 4.6976 mol initial moles of F_2 = 80.1 g/37.9968 g/mol = 2.1054 mol here F_2 is limiting reactant because required moles 0 % F_2 are more than moles of F_2 initially reactant F_2 from stoichiometres moles of H_2 O produced/moles of F_2 reacted = 2/5 moles of H_2 O produced = 2/5 times moles of F_2 reacted = 2/5 times 2.1054 = 0.84216 moles Theoretical yield = theoretical moles of H_2 O times H_2 O molar mass = 0.84216 mol times 18.0152 g/mol = 15.17 g H_2O percent yield = actual yield/Theoretical yield times 100 = 9.73 g/15.17 g times 100 = 64.14 %