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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)?

Respuesta :

Explanation:

Reaction for decomposition of [tex]H_{2}O_{2}[/tex] is as follows.

             [tex]2H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}[/tex]

Molarity of [tex]H_{2}O_{2}[/tex] = 2.57 M

Volume = 4.20 ml

             = [tex]4.20 ml \times \frac{0.001 L}{1 ml}[/tex]

             = 0.0042 L

Now, calculate the moles of [tex]H_{2}O_{2}[/tex] as follows.

             Moles of [tex]H_{2}O_{2}[/tex] = Molarity × Volume

                                           = [tex]2.57 M \times 0.0042 L[/tex]

                                           = 0.0107 mol

According to the balanced equation, 2 mole of [tex]H_{2}O_{2}[/tex] gives 1 mole of [tex]O_{2}[/tex].

Hence, 0.0107 mol of [tex]H_{2}O_{2}[/tex] gives 0.00575 mol of [tex]O_{2}[/tex].

Partial pressure of [tex]O_{2}[/tex] = 0.9624 atm

Temperature = 300.95 K

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

                              PV = nRT

                         V = [tex]\frac{nRT}{P}[/tex]

                            = [tex]\frac{0.00575 mol \times 0.0821 Latm/mol K \times 300.95 K}{0.9624 atm}[/tex]

                             = 0.1475 L

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

Thus, we can conclude that volume of [tex]O_{2}[/tex] is 147.5 ml.

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