Question

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

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

Answer 1

`"8.90g"`.

Explanation:

This is our balanced equation (I think you meant to write `2H_2O` instead of `2H2)`?):

`2H_2O -> 2H_2 + O_2`

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

First, let's find the number of moles of `H_2O` in `"10.0g"`:

`"number of moles" = "mass of sample"/"mass of 1 mole"`

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

`2xxH + O = 2xx1.008 + 16.00 = "18.02g"`

Let's plug that into our equation:

`"number of moles" = "10.0g"/"18.02g" = 0.555`

For `2` moles of water, we get `1` mole of `O_2`.
That means that, for `0.555` moles of water, we get `0.555/2 = 0.278` moles of `O_2`.

Now that we know the number of moles of `O_2`, we can find its mass by multiplying the number of moles by the mass of `1` mole.
We can find the mass of `1` mole of `O_2` using the same method as we did for `H_2O`:

`2xxO = 2xx16.00 = "32.00g"`
`"mass" = 0.278 xx "32.00g" = "8.90g"`