Question

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`?

Answer 1

`"mass"` (`"O"_2`)`= 32` `"g"`

Explanation:

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

`2"H"_2"O"_2 -> 2"H"_2"O" + "O"_2`

First, calculate the moles of `"H"_2"O"_2` reacting. In order to do this, we must evaluate the relative molecular mass (`"M"_r`) of hydrogen peroxide:

`"M"_r` (`"H"_2"O"_2`) `= 2xx1 + 2xx16 = 34`

`"mol" = "m"/"M"_r = 68/34 = 2` `"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:

`"mol"` (`"H"_2"O"_2`) : `"mol"` (`"O"_2`)
`" "2" ":" "1`

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

`:.` `"mol"` (`"O"_2`)`= ("mol"("H"_2"O"_2))/2 = 2/2 = 1` `"mole"`

Finally, convert from moles into mass, given that `"M"_r` (`"O"_2`)`= 32` :

`"mol" = "m"/"M"_r => "m" = "mol"xx"M"_r = 1 xx 32 = 32` `"g"`