20. The following is a Limiting Reactant problem:

Magnesium nitride is formed in the reaction of magnesium metal with nitrogen gas in this reaction: 3 Mg (s) + N2 (g) - -> Mg3N2 (s)

How many grams of product are formed from 2.0 mol of N2 (g) and 8.0 mol of Mg (s) ? Show all calculations leading to an answer.

201.90 grams.

Explanation

Based on the equation:

For each mole of Mg (s) consumed, 1/3 mole of Mg₃N₂ will be produced. For each mole of N₂ (g) consumed, 1 mole of Mg₃N₂ will be produced. 8/3 ≈ 2.67 moles of Mg₃N₂ will be produced if all 8 moles of Mg (s) are consumed. 2.0/1 = 2.0 moles of Mg₃N₂ will be produced if all 2 moles of N₂ (g) are consumed.

Only 2.0 moles of Mg₃N₂ will be produced in the end. The reaction will run out of N₂ (g) before all Mg (s) is consumed. N₂ (g) is the limiting reactant. As a result, the quantity of N₂ (g) supplied determines the quantity of Mg₃N₂ produced.

Refer to a periodic table for relative atomic mass values. Mg₃N₂ has a molar mass of 3 * 24.31 + 2 * 14.01 = 100.95 g/mol. As a result, the 2.0 mol of Mg₃N₂ produced will have a mass of 2 * 100.95 = 201.90 grams.