What is the maximum mass of ammonia that can be formed when 43.00 grams ....

Question

What is the maximum mass of ammonia that can be formed when 43.00 grams of nitrogen gas reacts with 10.62 grams of hydrogen gas according to the following equation? Please round your answer to the nearest 0.01 gram.(thank you in advance)!

* N₂ + 3 H₂ = 2 NH₃

Answer 1

So, any time you are given the amounts of BOTH reactants, the first thing you must do (after balancing the equation) is to find the limiting reactant. One way to do that is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less indicates the limiting reactant (see below for this problem).

N₂ + 3H₂ ==> 2NH₃ .. balanced equation

moles N₂ = 43.00 g x 1 mol N₂ / 28 g = 1.54 mols N₂ (÷1->1.54)

moles H₂ = 10.62 g x 1 mol H₂/ 2 g = 5.31 mols H₂ (÷3->1.77)

Since 1.54 is less than 1.77, this tells us that N₂ is the limiting reactant, and the moles of N₂(1.54 mols) will determine how much NH₃ can be formed.

Now, we use 1.54 mols of N₂ and the stoichiometry of the balanced equation, along with dimensional analysis to get our answer:

mass of NH₃ formed = 1.54 mols N₂ x 2 mols NH₃ / mol N₂ x 17 g NH₃ / mol NH₃ = 52.36 g NH₃