Question

What volume of O2 is produced when 28 g of hydrogen peroxide (H2O2) decomposes to form water and oxygen at 150°C and 2.0 atm?___ H2O2 (aq) → ___ H2O (l) + ___ O2 (g)

Answer 1

The volume of O2 is 7.11L.

Step-by-step explanation:

1st) It is necessary to balance the chemical reaction:

`2H_2O_2\rightarrow2H_2O+O_2`

The reaction is balanced, because there is the same amount of each element in both sides of the reaction.

2nd) From the balanced reaction, we know that from 2 moles of hydrogen peroxide (H2O2), 1 mole of oxygen gas is produced. Now, using the molar mass of hydrogen peroxide (34g/mol), we can calculate the number of moles of O2 that will be produced from 28g of H2O2:

- Conversion from grams to moles:

`28g*(1mole)/(34g)=0.82moles`

- Number of moles of O2 produced:

`\begin{gathered} 2molesH_2O_2-1moleO_2 \\ 0.82molesH_2O_2-x=(0.82molesH_2O_2*1moleO_2)/(2molesH_2O_2) \\ x=0.41molesO_2 \end{gathered}`

3rd) Finally, we can calculate the volume of O2 using the Ideal Gases formula, replacing the values of pressure (2.0atm), number of moles (0.41moles) and temperature (150°C=423K):

`\begin{gathered} P*V=n*R*T \\ 2.0atm*V=0.41moles*0.082(atm*L)/(mol*K)*423K \\ 2.0atm*V=14.22atm*L \\ V=(14.22atm*L)/(2.0atm) \\ V=7.11L \end{gathered}`

Remember that to use the Ideal Gases formula, the units must be: atm, liter, mol and Kelvin.

So, the volume of O2 is 7.11L.