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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?

 Mar 9, 2019
 #1
avatar+290 
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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 laugh

 Mar 10, 2019

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