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 $H_2O_2$?

Md Ariful Islam

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Md Ariful Islam

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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$ :

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

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