Potassium superoxide, KO2, is an ionic solid. It can be used in a spacecraft to supply oxygen according this equation:
4KO2 (s) + 2H2O (l) --> 4KOH (s) + 3O2 (g).
The formed potassium hydroxide removes carbon dioxide.
Show that 1.0g of potassium superoxide will supply about 0.25dm^3 of oxygen at room temp. and RTP.
MR of KO2 = 71.1
1/7.1 = 0.14mol of KO2.
0.14mol of KO2 produces 0.11mol of oxygen. (4:3 molar ratio)
then how ah?
Number of Moles = Sample Volume / Molar Volume
Sample Volume = No. of Moles x Molar Volume = 0.01055 x 24 dm3 = 0.253 dm3
after finding the no. of moles of O2 produced, i converted it to dm3, and it was 2.64..
but if you use sample vol(0.25) / molar vol (24), you would get 0.0104, which is not the case following the chemical equation..(0.11)
dont quite understand..
Originally posted by donkhead333:after finding the no. of moles of O2 produced, i converted it to dm3, and it was 2.64..
but if you use sample vol(0.25) / molar vol (24), you would get 0.0104, which is not the case following the chemical equation..(0.11)
dont quite understand..
The qn says VERIFY that sample vol is 0.25dm3, in other words, you're not supposed to know or use sample volume as 0.25dm3 at all. You're supposed to OBTAIN 0.25dm3, based on stoichiometry.