Chemistry percent yield?

Molar mass of C₂H₆ = (12.0×2 + 1.0×6) g/mol = 30.0 g/mol
Molar mass of O₂ = 16 × 2 g/mol = 32.0 g/mol
Molar mass of H₂O = (10×2 + 16) g/mol = 18.0 g/mol

Initial number of moles of C₂H₆ = (2.10 g) / (30.0 g/mol) = 0.0700 mol
Initial number of moles of O₂ = (14.1 g) / (32.0 g/mol) = 0.441 mol

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Mole ratio C₂H₆ : O₂ : H₂O = 2 : 7 : 6

If 0.0700 mol C₂H₆ completely burned, O₂ needed = 0.0700 × (7/2) = 0.245 mol < 0.441 mol
O₂ is in excess, and thus C₂H₆ is the limiting reactant/reagent.

Number of moles of C₂H₆ reacted = 0.0700 mol
Maximum number of moles of H₂O formed = (0.0700 mol) × (6/2) = 0.210 mol
Theoretical yield of H₂O = (0.210 mol) × (18.0 g/mol) = 3.78 g

Percent yield of H₂O = (0.981/3.78) × 100% = 26.0%

Let’s determine the number of moles of water.
n = 0.981 ÷ 18 = 0.0545

Reaction:

C2H6 + O2 → CO2 + H2O
To balance the C’s make 2 CO2. To balance the H’s, make 3 H2O

C2H6 + O2 → 2 CO2 + 3 H2O
To balance the O’s, make 3.5 O2.
C2H6 + 3.5 O2 → 2 CO2 + 3 H2O
To balance the equation, double all of the coefficients.
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

According to the coefficients, two moles of ethane reacts with seven moles of ethane to make four moles of carbon dioxide and six moles of water.

Ethane = C2H6
Mass of one mole = 24 + 6 = 30 grams
n = 2.10 ÷ 30 = 0.07
For O2, n = 14.1 ÷ 32 = 0.440625

Mole ratio = 0.440625 ÷ 0.07
This is approximately 6.3:1. This is greater than 3.5. This means we have excess oxygen. Let’s use the following proportion to determine the number of moles of water that should have been produced.

2 : 6 = 0.07 * x
x = 0.21
Now we can determine the percent yield.

% yield = 100 * 0.0545 ÷ 0.21
This is approximately 26%