Question

Two moles of oxygen gas is reacted with hydrogen gas to produce water in this reaction: 2 H2 + O2 → 2 H2O In the lab you actually made 33.1 grams of water. What is the percent yield? 218.10% 91.80% 45.87% 54.50%

Answer 1

Theoretical yield:

02 -> H2O
1mol -> 2mol
2mol -> 4mol
32g -> 72g

Theoretical yield = 72g

Percent yield = actual yield/theoretical yield * 100

= 33.1/72 *100

=45.97%

Answer 2

Answer : The percent yield of $H_2O$ is, 45.97 %

Explanation :

First we have to calculate the moles of water.

The balanced chemical reaction is,

$2H_2+O_2\rightarrow 2H_2O$

From the balanced reaction, we conclude that

As, 1 moles of $O_2$ react to give 2 moles of $H_2O$

So, 2 moles of $O_2$ react to give $2\times 2=4$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$.

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(4mole)\times (18g/mole)=72g$

The mass water produces, 72 g

Now we have to calculate the percent yield of $H_2O$.

$\%\text{ yield of }H_2O=\frac{\text{Actual yield of }H_2O}{\text{Theoretical yield of }H_2O}\times 100=\frac{33.1g}{72g}\times 100=45.97\%$

Therefore, the percent yield of $H_2O$ is, 45.97 %