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? Use mc022-2.jpg.

Answer 1

88.5% is the correct answer according to my calculations.

Answer 2

Answer :  The percent yield of $H_2O$ is, 87.88%

Solution :

First we have to calculate the moles of $H_2$ and $O_2$.

$\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{11g}{2g/mole}=5.5moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{95g}{32g/mole}=2.9moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2H_2+O_2\rightarrow 2H_2O$

From the balanced reaction we conclude that

As, 2 moles of $H_2$ react with 1 mole of $O_2$

So, 5.5 moles of $H_2$ react with $\frac{5.5}{2}=2.75$ moles of $O_2$

That means, in the given balanced reaction, $H_2$ is a limiting reagent because it limits the formation of products and $O_2$ is an excess reagent.

The excess reagent remains $(O_2)$  = 2.9 - 2.75 = 0.15 moles

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

As, 2 moles of $H_2$ react with 2 moles of $H_2O$

As, 5.5 moles of $H_2$ react with 5.5 moles of $H_2O$

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

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

$\text{Mass of }H_2O=(5.5mole)\times (18g/mole)=99g$

Therefore, the mass water produces, 99 g

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

$\%\text{ yield of }H_2O=\frac{\text{Actual yield of }H_2O}{\text{Theoretical yield of }H_2O}\times 100=\frac{87g}{99g}\times 100=87.88\%$

Therefore, the percent yield of $H_2O$ is, 87.88%