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}=(11g)/(2g/mole)=5.5moles`

`\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=(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
`(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* \text{Molar mass of }H_2O`

`\text{Mass of }H_2O=(5.5mole)* (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}* 100=(87g)/(99g)* 100=87.88\%`

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