Chemistry Stoich

Question

In the Haber process, 32.00 g of nitrogen gas is combined with 40.00 g of hydrogen gas to form ammonia gas, NH₃. What mass, in grams, of ammonia will form? 

Answer 1

First, write the correctly balanced equation for the reaction taking place:

N₂(g) + 3H₂(g) ==> 2NH₃(g) ... balanced equation

Next, since we are given the amounts of BOTH reacts, we must first find the limiting reactant. One easy way to do this is to divide the MOLES of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

For N₂: 32.00 g N₂ x 1 mol N₂ / 28.01 g = 1.142 moles N₂ (÷1->1.1)

For H₂: 40.00 g H₂ x 1 mol H₂ / 2.016 g = 19.84 moles H₂ (÷3->6.6)

Since 1.1 is less than 6.6, N₂ is the limiting reactant

Finally, we use the MOLES of the limiting reactant (1.142 mols N₂) to calculate mass of NH₃ formed. We do this using the stoichiometry of the balanced equation, the molar mass of NH₃ (17.03 g/mol), and dimensional analysis:

1.142 moles N₂ x 2 mols NH₃ / 1 mol N₂ x 17.03 g NH₃ / mol NH₃ = 38.90 g NH₃ (4 sig.figs.)