Question #4ea7f

1 Answer
Jan 4, 2016

Answer:

Here's what I got.

Explanation:

The trick here is to realize that the volume of oxygen gas collected over water will contain water vapor.

This implies that the actual pressure of oxygen gas will be lower than the measured #"756 mmHg"#, since that pressure will contain the partial pressure of the water vapor at that temperature.

At #23^@"C"#, water has a vapor pressure of about #"21.0 mmHg"#

http://www.endmemo.com/chem/vaporpressurewater.php

This means that the pressure of the oxygen gas will be

#P_"total" = P_"water" + P_"oxygen"#

#P_"oxygen" = "756 mmHg" - "21.0 mmHg" = "735 mmHg"#

Now, your next move will be to calculate the number of moles of oxygen gas produced by the decomposition reaction.

If you assume that your hydrogen peroxide solution contains #"98 mL"# of hydrogen peroxide and that its density is equal to #"1 g/mL"#, you can say that

#98 color(red)(cancel(color(black)("mL"))) * ("1 g H"_2"O"_2)/(1color(red)(cancel(color(black)("mL")))) = "98 g H"_2"O"_2#

Use hydrogen peroxide's molar mass to determine how many moles undergo decomposition

#98 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O"_2)/(34.015color(red)(cancel(color(black)("g")))) = "2.88 moles H"_2"O"_2#

Write a balanced chemical equation for this reaction

#color(red)(2)"H"_2"O"_text(2(aq]) -> 2"H"_2"O"_text((l]) + "O"_text(2(g]) uarr#

Notice the #color(red)(2):1# mole ratio that exists between hydrogen peroxide and oxygen gas. Use it to find the number of moles of oxygen gas produced by the reaction

#2.88 color(red)(cancel(color(black)("moles H"_2"O"_2))) * "1 mole O"_2/(color(red)(2)color(red)(cancel(color(black)("moles H"_2"O"_2)))) = "1.44 moles O"_2#

Now you're ready to use the ideal gas law equation to find the value of the universal gas constant, #R#

#color(blue)(PV = nRT implies R = (PV)/(nT))#

Do not forget to convert the pressure of the gas from mmHg to atm and the temperature from degrees Celsius to Kelvin!

#R = ( 735/760 "atm" * "72.5 L")/("1.44 moles" * (273.15 + 23)"K")#

#R = 0.1644("atm" * "L")/("mol" * "K")#

This result is off by about #100%#, since the actual value of #R# is

#R = 0.0821("atm" * "L")/("mol" * "K")#

My guess is that the amount of hydrogen peroxide given to you is wrong. You will get a great result for

#n_"oxygen" = "2.88 moles"#

which in turn would imply that

#n_(H_2O_2) = "5.76 moles"#

So you'd need about twice as much hydrogen peroxide to produce that many moles, and result in

#R = 0.0822("atm" * "L")/("mol" * "K")#

a value that is much, much closer to the accepted value.

Now, in order to have a density of #"1 g/mL"# at that temperature, you'd need a hydrogen peroxide solution that is below #"10% w/w"#. So if the volume of hydrogen peroxide is #"98 mL"#, then you can expect the solution to have a volume of about #"1.0 L"#.