Question

what mass of magnesium is required to react with excess nitrogen to produce 0.500 mol magnesium nitrade

Answer 1

We need 36.45 grams of Mg

Step-by-step explanation:

Step 1: Data given

Moles of Mg3N2 produced = 0.500 moles

Molar mass of Mg = 24.3 g/mol

Molar mass of Mg3N2 = 256.41 g/mol

Step 2: The balanced equation

3Mg + N2 → Mg3N2

Step 3: Calculate moles of Mg

For 3 moles of Mg we need 1 mol of N2 to produce 1 mol of Mg3N2

For 0.500 moles Mg3N2 we need 3*0.500 moles = 1.500 moles of Mg

Step 4: Calculate mass of Mg

Mass Mg = moles Mg * molar mass Mg

Mass Mg = 1.500 moles * 24.3 g/mol

Mass Mg = 36.45 grams

Answer 2

36.46 g

Step-by-step explanation:

Let's consider the following balanced equation.

3 Mg + N₂ → Mg₃N₂

The molar ratio of Mg to Mg₃N₂ is 3:1. The moles of Mg that produce 0.500 moles of Mg₃N₂ are:

0.500 mol Mg₃N₂ × (3 mol Mg/ 1 mol Mg₃N₂) = 1.500 mol

The molar mass of Mg is 24.305 g/mol. The mass corresponding to 1.500 moles is:

1.500 mol × (24.305 g/mol) = 36.46 g