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Oxygen gas, generated by the reaction 2KClO3 (s) 2KCl(s) 3O2 (g) is collected over water at 27°C in a 1.35-L vessel at a total pressure of 1.00 atm. (The vapor pressure of H2O at 27°C is 26.0 torr.) How many moles of KClO3 were consumed in the reaction?

Answer :

sebassandin

Answer:

[tex]n_{KClO_3}=0.0353molKClO_3[/tex]

Explanation:

Hello!

In this case, according to the experiment, we first realize that the pressure of oxygen is:

[tex]p_{O_2}=1.00atm-26.0/760 atm=0.966atm[/tex]

Next, we compute the moles based off the ideal gas law:

[tex]PV=nRT\\\\n=\frac{PV}{RT}=\frac{0.966atm*1.35L}{0.08206\frac{atm*L}{mol*K}*300K}=0.0530molO_2[/tex]

Thus, since there is 3:2 mole ratio between oxygen and potassium chlorate, we can compute the consumed moles of the latter as shown below:

[tex]n_{KClO_3}=0.0530molO_2*\frac{2molKClO_3}{3molO_2} \\\\n_{KClO_3}=0.0353molKClO_3[/tex]

Best regards!

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