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* 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* \text{Molar mass of }H_2O`

`\text{Mass of }H_2O=(4mole)* (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}* 100=(33.1g)/(72g)* 100=45.97\%`

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