Question #4ea7f
1 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
At
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 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
#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,
#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
#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