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
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Ответ:
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
Ответ: