Question

For the balanced equation shown below, if the reaction of 40.8 grams of C6H6O3 produces a 39.0% yield, what is the theoretical yield and how many grams of H2O would be produced?

(C = 12.01 amu, H = 1.01 amu, O = 16.00 amu)

C6H6O3 + 6 O2 → 6 CO2 + 3 H2O

Theoretical yield = Blank 1 grams of H2O

Actual yield = Blank 2 grams of H2O

Answer 1

1)

Molar mass of C6H6O3 = 6*MM(C) + 6*MM(H) + 3*MM(O)

= 6*12.01 + 6*1.01 + 3*16.0

= 126.11 g/mol

mass of C6H6O3 = 40.8 g

mol of C6H6O3 = (mass)/(molar mass)

= 40.8/126.11

= 0.3235 mol

From balanced chemical reaction, we see that

when 1 mol of C6H6O3 reacts, 3 mol of H2O is formed

mol of H2O formed = (3/1)* moles of C6H6O3

= (3/1)*0.3235

= 0.9706 mol

Molar mass of H2O = 2*MM(H) + 1*MM(O)

= 2*1.01 + 1*16.0

= 18.02 g/mol

mass of H2O = number of mol * molar mass

= 0.9706*18.02

= 17.5 g

2)

% yield = actual mass*100/theoretical mass

39.0= actual mass*100/17.5

actual mass=6.82 g

Theoretical yield:17.5 g

Actual yield : 6.82 g