Question

Ozzie wanted to do another experiment with a stronger H₂O₂ solution to check the accuracy of the experiment by calculating the theoretical volume of O₂(g) it would produce. Then he could compare his experimental volume of O₂(g) to the theoretical volume of O₂(g). He used 4.20 mL of 2.57 M H₂O₂ and the partial pressure of O₂ was 0.9624 atm and the temperature was 300.95 K. What volume of O₂(g) could he theoretically produce (in mL)?

Answer 1

Step-by-step explanation:

Reaction for decomposition of
`H_(2)O_(2)` is as follows.

`2H_(2)O_(2) \rightarrow 2H_(2)O + O_(2)`

Molarity of
`H_(2)O_(2)` = 2.57 M

Volume = 4.20 ml

=
`4.20 ml * (0.001 L)/(1 ml)`

= 0.0042 L

Now, calculate the moles of
`H_(2)O_(2)` as follows.

Moles of
`H_(2)O_(2)` = Molarity × Volume

=
`2.57 M * 0.0042 L`

= 0.0107 mol

According to the balanced equation, 2 mole of
`H_(2)O_(2)` gives 1 mole of
`O_(2)`.

Hence, 0.0107 mol of
`H_(2)O_(2)` gives 0.00575 mol of
`O_(2)`.

Partial pressure of
`O_(2)` = 0.9624 atm

Temperature = 300.95 K

Now, using ideal gas equation we will calculate the volume as follows.

PV = nRT

V =
`(nRT)/(P)`

=
`(0.00575 mol * 0.0821 Latm/mol K * 300.95 K)/(0.9624 atm)`

= 0.1475 L

or, = 147.5 ml (as 1 L = 1000 mL)

Thus, we can conclude that volume of
`O_(2)` is 147.5 ml.