The actual yield of a product in a reaction was measured as 4.20 g. If the theoretical yield of the product for the reaction is 4.88 g, what is the percentage yield of the product?

86%

Explanation:

86.1%

Explanation:

The percentage yield of a product is the ratio of (actual yield over theoretical yield) multiplied by 100.

In this example, the actual yield is 4.2, and the theoretical yield is 4.88, which means the ratio is:

[4.2/4.88] *100 =86.1

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.