How many grams of o2(g) are needed to completely burn 46.0 g c3h8(g)?

167,3 grams of O2 (g)

Explanation:

According to the following balanced equation:

C3H8 + 5 02 ---> 3 CO2 + 4 H20

We calculate the mass of the moles:

5 mol O2 = 16grams/mol x2x 5= 160 grams/mol

1 mol C3H8= 3 x12 grams/mol + 8 x 1 gram/mol= 44 grams/mol

If 44 grams C3H8 react with 160 grams O2

46 gramsC3H8 react with    X = (46x160)/44 =167, 3 grams O2

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.