+687 A gas mixture contains 88 grams of oxygen gas, 88 grams of nitrogen gas, and 12 grams of nitrogen dioxide gas. What is the mole fraction of nitrogen dioxide in this mixture?
+368 I used the https://ptable.com
O2=2(15.999)=31.998
N2=2(14.007)=28.014
NO2=14.007+2(15.999)=46.005
Lets solve one by one using unit conversions
O2=1 mol/31.998 g
88g of O2= 88*1 mol/31.998g
88/31.998 moles of O2=2.531408213013313332083 moles of O2
N2=1 mol/28.014
88g of N2=88*1 mol/28.014
88/28.014 moles of N2=3.141286499607339187549 moles of N2
NO2= 1 mol/46.005
12g of NO2=12*1 mol/46.005
12/46.005 moles of NO2=0.2608412129116400391262 moles of NO2
0.2608412129116400391262/5.933535925532292558758=0.04396050115568149888328. To make life simpler, im going to round it to
0.04396.
Fractional form is
1099/25000
I dont know if this is correct-It doesnt seem like it-but I tried 
