Question

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.

Answer 1

Answers:

2.0 × 10² g Mg₃N₂

Step-by-step explanation:

We are given the masses of two reactants, so this is a limiting reactant problem.

We know that we will need moles and molar masses, so, lets assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r: 100.93

3Mg + N₂ ⟶ Mg₃N₂

n/mol: 8.0 2.0

Step 1. Identify the limiting reactant

Calculate the moles of Mg₃N₂ we can obtain from each reactant.

From Mg:

The molar ratio of Mg₃N₂:Mg is 1:3

Moles of Mg₃N₂ = 8.0 × 1/3

Moles of Mg₃N₂ = 2.67 mol Mg₃N₂

From N₂:

The molar ratio of Mg₃N₂: N₂ is 1:1.

Moles of Mg₃N₂ = 2.0 × 1/1

Moles of Mg₃N₂ = 2.00 mol Mg₃N₂

N₂ is the limiting reactant because it gives the smaller amount of Mg₃N₂.

Step 2. Calculate the theoretical yield.

Theor. yield = 2.00 mol Mg₃N₂ × 100.93 g Mg₃N₂/1 mol Mg₃N₂

Theor. yield = 2.0 × 10² g Mg₃N₂

Answer 2

201.90 grams.

Step-by-step 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.