Question

If 62.7 g N2 react with 23.8 g H2 according to the reaction below, how many grams of ammonia (NH3) can be produced, and how many grams of the excess reactant will be left over?

Unbalanced equation: N2 + H2 → NH3

Answer 1

How many moles of N2 and H2 do you have?

Answer 2

62.7 g N2 * mol/28g = 2.24 mol
23.8 g H2 * mol/2g= 11.9 mol ?????

Answer 3

Ok, now what molar ratio does the balanced reaction require? How many moles of H2 will react with 1mol of N2?\[N_{2} + H_{2} \rightarrow NH_{3}\]\[\frac {mol N_{2}}{mol H_{2}} = ?\]

Answer 4

N2 and H2 they both have 1:1 molar ratio

Answer 5

The reaction's not balanced, the ratio has to be: \[N_{2} + 3H_{2} \rightarrow 2NH_{3}\] 1 mole of N2 requires 3 moles of H2 in order to fully react. You have 2.24mol of N2 available, so how much H2 is needed to fully react with that, if 1mol of N2 requires 3 mol of H2?

Answer 6

Yeah I balanced it to- N2 + 3H2 = 2NH3

Answer 7

since 1 mole of N2 needs 3 moles of H2, how much H2 do you need if you have 2.24moles of N2?

Answer 8

So wait, how do I find H2?

Answer 9

the molar ratio is 1N_2 : 3H_2. 2.24 moles of N2 will require three times that much H2 because the balanced reaction says so

Answer 10

So do I multiply 2.24 by three?

Answer 11

H2 has 11.9 mol???

Answer 12

you have 11.9mol of H2, but only need 3*2.24mol of H2, so the N2 will run out first. The N2 is the "limiting reactant". The limiting reactant dictates how much product you can make.

The limiting reactant will be used up entirely, so how much NH3 will be made if you use all 2.24mol of N2?