Question

The following is a Limiting Reactant problem:

Magnesium nitride is formed in the reaction of magnesium metal with nitrogen gas in this reaction: 3 Mg(s) + N2(g) -----> Mg3N2(s)
How many grams of product are formed from 2.0 mol of N2 (g) and 8.0 mol of Mg(s)?

Answer 1

`3 Mg_((s)) + N _(2) _((g)) ----\ \textgreater \ Mg_(3) N _(2) _((s))`

mole ratio of Mg : Mg₃N₂ = 3 b: 1

∴ mole of Mg₃N₂ =
`(mole of magnesium)/(3)`

=
`(8.0 mol)/(3)`

= 2.67 mol

mass of Mg₃N₂ = mole * molar mass
= 2.67 mol * ((3 * 24) + (2 * 14)) g / mol
= 266.67 g

Answer 2

Answer is: 201.86 grams of product are formed.

Balanced chemical rection: 3Mg(s) + N₂(g) → Mg₃N₂(s).

n(N₂) = 2.0 mol; amount of nitrogen.

n(Mg) = 8.0 mol: amount og magnesium.

From balanced reaction: n(Mg) : n(N₂) = 3 : 1.

For 8 moles of magnesium:

8 mol : n(N₂) = 3 : 1.

n(N₂) = 2.66 mol.

There is not enough nitrogen to react completely with magnesiu, so nitrogen is limiting reactant.

From balanced reaction: n(N₂) : n(Mg₃N₂) = 1 : 1.

n(Mg₃N₂) = 2 mol.

m(Mg₃N₂) = n(Mg₃N₂) · M(Mg₃N₂).

m(Mg₃N₂) = 2 mol · 100.93 g/mol.

m(Mg₃N₂) = 201.86 g; mass of magnesium nitride.