The decomposition of $H_2O_2$ produces water and oxygen gas, releasing 197 kJ per one mole of $H_2O_2$ . How much energy is released if we start with 798 grams of $H_2O_2$ ?

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Answer 1 · 2016-05-24T12:38:49

$q=4629.5kJ$

The amount of heat ( $q$ ) released from decomposing $798g$ of $H_2O_2$ could be found by:

$q=DeltaHxxn$ where, $DeltaH$ is the enthalpy of the reaction and $n$ is the number of mole of $H_2O_2$ .

Note that $DeltaH=197kJ*mol^(-1)$

To find $n$ , we can simply use: $n=m/(MM)$ where, $m=798g$ is the given mass and $MM=34g*mol^(-1)$ is the molar mass of $H_2O_2$ .

$n=m/(MM)=(798cancel(g))/(34cancel(g)*mol^(-1))=23.5molH_2O_2$

Thus, $q=DeltaHxxn=197(kJ)/(cancel(mol))xx23.5cancel(mol)=4629.5kJ$

The decomposition of H_2O_2H_2O_2 produces water and oxygen gas, releasing 197 kJ per one mole of H_2O_2H_2O_2. How much energy is released if we start with 798 grams of H_2O_2H_2O_2?