How many grams of oxygen are produced if you have 10.0 grams of water available for the following reaction? 2H2)-> 2H2 + O2

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How many grams of oxygen are produced if you have 10.0 grams of water available for the following reaction? 2H2)-> 2H2 + O2

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Answer 1 · 2018-02-15T03:57:20

$8.90g$ $"8.90g"$ .

Explanation:

This is our balanced equation (I think you meant to write $2 H_{2} O$ $2H_2O$ instead of $2 H 2)$ $2H2)$ ?):

$2 H_{2} O \to 2 H_{2} + O_{2}$ $2H_2O -> 2H_2 + O_2$

From this balanced equation, we know that for every $2$ $2$ moles of water, there will be $1$ $1$ mole of oxygen gas, or $O_{2}$ $O_2$ .

First, let's find the number of moles of $H_{2} O$ $H_2O$ in $10.0g$ $"10.0g"$ :

$number of moles = \frac{mass of sample}{mass of 1 mole}$ $"number of moles" = "mass of sample"/"mass of 1 mole"$

We know the mass of the sample ( $10.0g$ $"10.0g"$ ), but we're still missing the mass of $1$ $1$ mole of water—but, we can find that using the masses of $2$ $2$ moles of hydrogen and $1$ $1$ mole of oxygen!:

$2 \times H + O = 2 \times 1.008 + 16.00 = 18.02g$ $2xxH + O = 2xx1.008 + 16.00 = "18.02g"$

Let's plug that into our equation:

$number of moles = \frac{10.0g}{18.02g} = 0.555$ $"number of moles" = "10.0g"/"18.02g" = 0.555$

For $2$ $2$ moles of water, we get $1$ $1$ mole of $O_{2}$ $O_2$ .
That means that, for $0.555$ $0.555$ moles of water, we get $\frac{0.555}{2} = 0.278$ $0.555/2 = 0.278$ moles of $O_{2}$ $O_2$ .

Now that we know the number of moles of $O_{2}$ $O_2$ , we can find its mass by multiplying the number of moles by the mass of $1$ $1$ mole.
We can find the mass of $1$ $1$ mole of $O_{2}$ $O_2$ using the same method as we did for $H_{2} O$ $H_2O$ :

$2 \times O = 2 \times 16.00 = 32.00g$ $2xxO = 2xx16.00 = "32.00g"$
$mass = 0.278 \times 32.00g = 8.90g$ $"mass" = 0.278 xx "32.00g" = "8.90g"$

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How many grams of oxygen are produced if you have 10.0 grams of water available for the following reaction? 2H2)-> 2H2 + O2