Question #0cba7

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Question #0cba7

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Answer 1 · 2017-07-15T10:51:23

The yield will be $2$ $2$ $m o l$ $mol$ of $H_{2} O$ $H_2O$ .

Explanation:

First we need to write a balanced chemical equation:

$2 H_{2} + O_{2} = 2 H_{2} O$ $2H_2+O_2=2H_2O$

In this case, the $O_{2}$ $O_2$ will be the limiting reagent: we need $2$ $2$ $m o l$ $mol$ of $H_{2}$ $H_2$ for each mole of $O_{2}$ $O_2$ , but we have 4. There will be some $H_{2}$ $H_2$ left over unreacted.

We know from the balanced equation that $1$ $1$ $m o l$ $mol$ of $O_{2}$ $O_2$ will yield $2$ $2$ $m o l$ $mol$ of $H_{2} O$ $H_2O$ , and since we have $1$ $1$ $m o l$ $mol$ of $O_{2}$ $O_2$ the yield will be $2$ $2$ $m o l$ $mol$ of $H_{2} O$ $H_2O$ .

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Question #0cba7

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