We’re being asked to calculate the **percent yield of H _{2}O. **The balanced chemical equation is:

**2 H _{2} + O_{2} → 2 H_{2}O**

Recall that ** percent yield** is given by:

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{Y}}{\mathbf{i}}{\mathbf{e}}{\mathbf{l}}{\mathbf{d}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{u}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{y}\mathbf{i}\mathbf{e}\mathbf{l}\mathbf{d}}{\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{o}\mathbf{r}\mathbf{e}\mathbf{t}\mathbf{i}\mathbf{c}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{y}\mathbf{i}\mathbf{e}\mathbf{l}\mathbf{d}}{\mathbf{}}{\mathbf{\times}}{\mathbf{}}{\mathbf{100}}}$

In the reaction of 1.00 mole of O_{2} with 2.00 mole of H_{2}, 2H_{2} + O_{2}→ 2H_{2}O

If 30.0g of H_{2}O is formed, what is the percent yield?

a) 100%

b) 88.9%

c) 50.0%

d) 167%

e) 83.3%

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Percent Yield concept. If you need more Percent Yield practice, you can also practice Percent Yield practice problems.