Answer to Question #125158 in General Chemistry for mike

Question

What is the limiting reactant when 1.50g of magnesium and 1.50g of nitrogen combine according to the equation:
3Mg(s) + N2(g) ⇨ Mg3N2(s)
and how many grams of Mg3N2 are formed?

Answer 1

3Mg(s) + N₂(g) $\implies$ Mg₃N₂(s)

Moles of Mg = 1.5/24

Moles of N₂ = 1.5/28

For finding limiting reagent we will divide no. of moles by stoichiometric coefficient in balance reaction and find which value is less that will be the limiting reagent.

So on solving we find that Mg will be the limiting reagent due to less value.

Now after finding reagent apply mole concept.

Mole concept is always based on limiting reagent.

Hence the no. of moles of Mg₃N₂ will formed is calculated via unitary method.

Unitary method

3 mole Mg will give 1 mole Mg₃N₂

So 1.5/24 moles will give 1.5/24*3 moles Mg₃N₂

Hence mass of Mg₃N₂ = (1.5/72)*100

Since molecular mass of Mg₃N2₂ is 100 gm

Mass of Mg₃N₂ formed = 2.0833 gm.

This is the answer.

Mass = moles * molecular mass

Stoichiometric coefficient are no. of moles used in a balance reaction.