Given a balance chemical equation,N2+3H2⟶2NH3 If a reaction mixture contains 2 moles of N2 and 7 moles of H2, what is the limiting reactant and what is the mass(g)remaining of the reactant in excess?

Question

Given a balance chemical equation,

      N2(g) + 3 H2(g)  2 NH3(g)

If a reaction mixture contains 2 moles of N2(g) and 7 moles of H2(g), which reactant is the limiting reactant, and what is the mass (in grams) remaining of the reactant in excess? 

Answer 1

N₂(g) + 3H₂(g) → 2NH₃(g) is already balanced. Notice there are 2 nitrogens on the left and 2 on the right. There are 6 hydrogens on the left and 6 on the right.

Using that balance chemical reaction equation, it says that for every 1 mole of nitrogen gas, N₂(g), we need 3 moles of hydrogen gas, H₂(g) so, since we are given 2 moles of N₂(g) we will need 6 moles of H₂(g). But we are given 7. That means we have extra hydrogen gas H₂(g) or, in other words, nitrogen gas N₂(g) is the limiting reactant. Without more N₂(g) we will not be able to use 1 of the moles of H₂(g) in the reaction.

To know the mass in grams of the 1 mole of hydrogen gas, look up the atomic mass of hydrogen on the periodic table and multiply it by 2 (there are 2 atoms of hydrogen in H₂). That will give you the molar mass (units will be grams/mole). Then multiply that result by the number of excess moles of hydrogen (1) to get the mass of the excess hydrogen in grams.