Balance the following redox equation in acidic solution using the smallest integers possible and select the correct coefficient for the h+(aq) ion.
cr2o72–(aq) + sn2+(aq) → cr3+(aq) + sn4+(aq)
(a) 1 (no coefficient written)
(b) 2
(c) 3
(d) 4
(e) more than 4

Coefficient of is more than 4

Explanation:

Oxidation:

Reduction:

gives balanced equation:

So coefficient of is more than 4

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.