Question

Consider the chemical equation. 2H2 + O2 mc022-1.jpg 2H2O What is the percent yield of H2O if 87.0 g of H2O is produced by combining 95.0 g of O2 and 11.0 g of H2?

Answer 1

The correct answer is letter c 88.5%. The percent yield of H2O if 87.0 g of H2O is produced by combining 95.0 g of O2 and 11.0g of H2 is 88.5%

Here are the choices.
a 56.5% 
b 59.0% 
c 88.5% 
d 99.7%

Answer 2

Answer:

88.50 %

The balance chemical equation is as follow,

2 H₂ + O₂ → 2 H₂O

Step 1: Find the limiting reactant;

According to eq.

4.032 g (2 mole) H₂ reacts with = 32 g (1 mole) of O₂

So,

11 g of H₂ will react with = X g of O₂

Solving for X,

X = (11 g × 32 g) ÷ 4.032 g

X = 87.301 g of O₂

Therefore, H₂ is the limiting reactant as O₂ is present in excess.

Step 2: Calculating %age Yield;

According to eq.

4.032 g (2 mole) H₂ produces = 36.032 g (1 mole) of H₂O

So,

11 g of H₂ will react with = X g of H₂O

Solving for X,

X = (11 g × 36.032 g) ÷ 4.032 g

X = 98.301 g of H₂O

So,

Actual Yield = 87 g

Theoretical Yield = 98.301 g

Using formula = Actual Yield / Theoretical Yield × 100

= 87 g / 98.301 × 100

= 88.50 %