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

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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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