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%

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