a solution contains 10.0 g pentane, c5h12, 20.0 g hexane, c6h14, and 10.0 g benzene, c6h6. what is the mole fraction of hexane?
a) 0.303
b) 0.471
c) 0.433
d) 0.362
e) 0.335

b) 0.47

Explanation:

MwC5H12 = 72.15g/mol

⇒mol C5H12 = (10.0)*(mol/72.15)=0.1386molC5H12

MwC6H14=86.18g/mol

⇒molC6H14=(20.0)*(mol/86.18)=0,232

MwC6H6=78.11g/mol

⇒molC6H6=(10.0)*(mol/78.11)=0.128molC6H6

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

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