Question

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

Answer 1

The correct answer is a. 2.0 moles of magnesium nitride are formed when 6.0 moles of magnesium react with 4.0 moles of nitrogen, leaving 2.0 moles of nitrogen unreacted as excess.

Step-by-step explanation:

The formation of magnesium nitride involves a reaction between magnesium metal and nitrogen gas. Magnesium forms a Mg2+ ion, which is a cation, by losing two electrons. Nitrogen, a non-metal, forms an anion with a charge of 3-. Therefore, the formula for magnesium nitride is Mg3N2, indicating that three magnesium ions react with two nitrogen ions to form one formula unit of magnesium nitride.

The balanced chemical equation for the formation of magnesium nitride is:

3Mg + N2 → Mg3N2

Given that 4.0 moles of nitrogen is reacted with 6.0 moles of magnesium, we can determine the limiting reactant and calculate the product formed. The molar ratio of Mg to N2 in the equation is 3:1, meaning three moles of Mg are required for every mole of N2. Therefore, 6.0 moles of Mg would completely react with 2.0 moles of N2 to produce magnesium nitride. Since we started with 4.0 moles of nitrogen, we have an excess of nitrogen, and magnesium is the limiting reactant.

The result is that all of the magnesium reacts and you would produce 2.0 moles of magnesium nitride with 2.0 moles of nitrogen left over as excess. Hence, the correct answer to the student's question is option a.

Answer 2

The correct answer is A

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

Step-by-step 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