Using the equation shown below (check for balance!), how many grams of ammonia will be formed if 75 grams of nitrogen reacts with excess hydrogen?

Question

Using the equation shown below (check for balance!), how many grams of ammonia will be formed if 75 grams of nitrogen reacts with excess hydrogen?

H₂ + N₂  NH₃

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Answer 1

First you need to balance the chemical reaction equation (notice the given hint).

1. Add coefficients in front of each chemical formula, so that there is the same number of atoms of each element on either side of the reaction.
2. Notice that, unbalanced, there are 2 hydrogen atoms on the reactant side, but 3 hydrogen atoms in the product. By adding a 3 in front of H₂ and a 2 in front of NH₃ you can "balance" the hydrogen atoms with 6 on both sides on the equation.
3. The you can balance the nitrogen atoms. Since we are adding the 2 in front of NH₃ we want to have 2 nitrogen atoms in the product. Good news! There are already 2 nitrogen atoms in the reactants. So we do not need to add any further coefficients.
4. The balanced equation is: 3 H₂ + N₂ → 2 NH₃

Next, you need to consider the second half of the question: "how many grams of ammonia will be formed if 75 grams of nitrogen reacts with excess hydrogen?"

1. This is a stoichiometry problem. We want to use the balanced reaction equation and the molar masses of ammonia and nitrogen, to "convert" 75 grams N₂ into a mass of NH₃.

3 "steps" for Mass to Mass Conversions:

1. Divide the given mass by the molar mass of the element/compound.
2. Multiply by the Mole Ratio (coefficients from the balanced equation).
3. Multiply by the molar mass of the element/compound you are asked to find.

Set everything up and always check to be sure your units cancel!

75 g N₂ x (1 mole N₂ / 28.014 g N₂) x (2 mole NH₃ / 1 mole N₂) x (17.025 g NH₃ / 1 mole NH₃) = 91 g NH₃