Question

9. Hydrogen peroxide decomposes to form water and oxygen gas according to the following equation:

2H2O2(aq) to 2H2O(l) + O2(g)
If 315 g of hydrogen peroxide, H2O2, decomposes and all the O2 gas is collected in a balloon at 0.792 atm and 23 degrees C, what is the volume of the O2 gas collected?

Answer 1

`V=142L`

Step-by-step explanation:

Hello,

In this case, for the given reaction:

`2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)`

Starting with 315 g of hydrogen peroxide, we can compute the yielded moles of oxygen by using the following stoichiometric factor whereas the hydrogen peroxide to oxygen mole ratio is 2:1:

`n_(O_2)=315gH_2O_2*(1molH_2O_2)/(34gH_2O_2)*(1molO_2)/(2molH_2O_2) \\\\n_(O_2)=4.63molO_2`

Then, by using the ideal gas equation we can compute the resulting volume if the 4.63 moles of oxygen are collected at 0.792 atm and 23 °C as shown below:

`PV=nRT\\\\V=(nRT)/(P)=(4.63mol*0.082(atm*L)/(mol*K)*(23+273.15)K)/(0.792 atm)\\ \\V=142L`

Best regards.