For the following reaction, 24.6 grams of nitrogen monoxide are allowed to react with 1.42 grams of hydrogen gas.

Question

nitrogen monoxide + hydrogen -> nitrogen + water

What is the maximum amount of nitrogen gas that can be formed?

Mass =  g

What is the FORMULA for the limiting reactant?

What amount of the excess reactant remains after the reaction is complete?

Mass =  g

Answer 1

Nitrogen monoxide = NO molar mass = 30 g / mol

Hydrogen = H₂ molar mass = 2 g / mol

Nitrogen= N₂ molar mass = 28 g / mol

water = H₂O molar mass = 18 g / mol

2NO + 2H₂ ==> N₂ + 2H₂O ... balanced equation

First we need to find the limiting reactant. An easy way is to divide the moles of each reactant by the corresponding coefficient in the balanced equation:

For NO we have 24.6 g x 1 mol / 30 g = 0.82 mols NO (÷2->0.41)

For H₂ we have 1.42 g x 1 mol / 2 g = 0.71 mols H₂ (÷2->0.35)

Since 0.35 is less than 0.41, this tells us the H₂ is limiting. We will now use the mols of H₂ (0.71 moles) to find the theoretical yield of N₂ gas.

0.71 mol H₂ x 1 mol N₂ / 2 mol H₂ x 28 g N₂ / mol N₂ = g N₂ that can be formed

To find mass of excess reagent (NO), we will find out how much NO was used in the reaction, and then subtract that from the amount we started with.

0.71 mol H₂ x 2 mol NO / 2 mol H₂ x 30 g NO / mol = 21.3 g NO used during the reaction

24.6 g - 21.3 g = 3.30 g NO remains after the reaction