Magnesium nitride is formed in the reaction of magnesium metal and nitrogen gas. 4.0 mol of nitrogen is reacted with 6.0 mol of magnesium. The result is ...

a. 2.0 mol of magnesium nitride and 2.0 mol excess nitrogen

b. 4.0 mol of magnesium nitride and 2.0 mol excess magnesium

c. 6.0 mol of magnesium nitride and 3.0 mol excess nitrogen

d. No product because the reactants are not in the correct mole ratio

The correct answer is A

We'll have 2.0 moles of magnesium nitride produced and and excess of 2.0 moles nitrogen gas

Explanation:

Step 1: Data given

Number of moles nitrogen gas (N2) = 4.0 moles

Number of moles magnesium = 6.0 moles

Step 2: The balanced equation

3Mg + N2 → Mg3N2

Step 3: Calculate the limiting reactant

For 3 molesmagnesium (Mg) we need 1 mol nitrogen gas (N2) to react, to produce 1 mol magnesium nitride (Mg3N2)

Magnesium is the limiting reactant. It will completely be consumed 6.0 moles. Nitrogen is in excess. There will react 6.0 / 3 = 2.0 moles

There will remain 4.0 - 2.0 = 2.0 moles N2

Step4: Calculate moles Mg3N2

For 3 molesmagnesium (Mg) we need 1 mol nitrogen gas (N2) to react, to produce 1 mol magnesium nitride (Mg3N2)

For 6.0 moles Mg we'll have 6.0 / 3 = 2.0 moles Mg3N2

The correct answer is A

We'll have 2.0 moles of magnesium nitride produced and and excess of 2.0 moles nitrogen gas