9. Hydrogen peroxide decomposes to form water and oxygen gas according to the following equation:
2H2O2(aq) to 2H2O(l) + O2(g)
If 315 g of hydrogen peroxide, H2O2, decomposes and all the O2 gas is collected in a balloon at 0.792 atm and 23 degrees C, what is the volume of the O2 gas collected?

Starting with 315 g of hydrogen peroxide, we can compute the yielded moles of oxygen by using the following stoichiometric factor whereas the hydrogen peroxide to oxygen mole ratio is 2:1:

[tex]n_{O_2}=315gH_2O_2*\frac{1molH_2O_2}{34gH_2O_2}*\frac{1molO_2}{2molH_2O_2} \\\\n_{O_2}=4.63molO_2[/tex]

Then, by using the ideal gas equation we can compute the resulting volume if the 4.63 moles of oxygen are collected at 0.792 atm and 23 °C as shown below:

[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{4.63mol*0.082\frac{atm*L}{mol*K}*(23+273.15)K}{0.792 atm}\\ \\V=142L[/tex]

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9. Hydrogen peroxide decomposes to form water and oxygen gas according to the following equation: 2H2O2(aq) to 2H2O(l) + O2(g) If 315 g of hydrogen peroxide, H2O2, decomposes and all the O2 gas is collected in a balloon at 0.792 atm and 23 degrees C, what is the volume of the O2 gas collected?

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9. Hydrogen peroxide decomposes to form water and oxygen gas according to the following equation:
2H2O2(aq) to 2H2O(l) + O2(g)
If 315 g of hydrogen peroxide, H2O2, decomposes and all the O2 gas is collected in a balloon at 0.792 atm and 23 degrees C, what is the volume of the O2 gas collected?