# In the formula H_2O_2 -> H_2O + O_2, how many grams of O_2 are produced from the decomposition of 68 g of H_2O_2?

Jan 4, 2016

$\text{mass}$ (${\text{O}}_{2}$)$= 32$ $\text{g}$

#### Explanation:

Your equation is not correctly balanced. The correct equation is as follows:

$2 {\text{H"_2"O"_2 -> 2"H"_2"O" + "O}}_{2}$

First, calculate the moles of ${\text{H"_2"O}}_{2}$ reacting. In order to do this, we must evaluate the relative molecular mass (${\text{M}}_{r}$) of hydrogen peroxide:

${\text{M}}_{r}$ (${\text{H"_2"O}}_{2}$) $= 2 \times 1 + 2 \times 16 = 34$

${\text{mol" = "m"/"M}}_{r} = \frac{68}{34} = 2$ $\text{moles}$

Next, we must compare the moles of hydrogen peroxide and oxygen gas. According to the corrected equation, this mole ratio is as shown below:

$\text{mol}$ (${\text{H"_2"O}}_{2}$) : $\text{mol}$ (${\text{O}}_{2}$)
$\text{ "2" ":" } 1$

Thus, the moles of oxygen produced in this reaction will be exactly half the moles of hydrogen peroxide reacting.

$\therefore$ $\text{mol}$ (${\text{O}}_{2}$) = ("mol"("H"_2"O"_2))/2 = 2/2 = 1 $\text{mole}$

Finally, convert from moles into mass, given that ${\text{M}}_{r}$ (${\text{O}}_{2}$)$= 32$ :

${\text{mol" = "m"/"M"_r => "m" = "mol"xx"M}}_{r} = 1 \times 32 = 32$ $\text{g}$