Explanation:
[tex]2H_2O_2 \rightarrow 2H_2O + O_2[/tex]
First convert the amount of water into moles:
360 g H2O × [tex]\left(\dfrac{1\:\text{mol}H_2O}{18.015\:\text{g}H_2O}\right)[/tex]
[tex] = 20. \:\text{mol}H_2O[/tex]
Now let's calculate the number of moles of O2 gas produced.
20 mol H2O × [tex]\left(\dfrac{1\:\text{mol}O_2}{2\:\text{mol}H_2O}\right)=10\:\text{mol}O_2[/tex]
The volume of gas at 10°C and 5 atm can be found using the ideal gas law:
[tex]PV=nRT[/tex]
[tex]V= \dfrac{nRT}{P}[/tex]
[tex]= \dfrac{(10)(0.082)(283)}{(5)}=46.4\:L[/tex]
