Question

20. 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)? Show all calculations leading to an answer

Answer 1

Can you help with this one @Photon336

Answer 2

or @aaronq

Answer 3

@Missiey

Answer 4

This the only equation we are gonna use here okai ?
no. of moles = mass/ molar mass
\[n = \frac{ m }{ M }\]
As we can see the molar ratios of Mg:N is 3:1
That means 3 moles of magnesium will react with 1 mole of nitrogen.
When u have 2 moles of nitrogen that means it will react with 6 (2*3) moles of magnesium.
But we have 8 moles of magnesium. So that means we have 2 moles (8-6) moles in excess.

Now look at the stoichiometric number of magnesium nitride. It is 1 .
Look at the molar ratio between nitrogen and magnesium nitride that is 1:1
That means one mole of nitrogen reacts to form 1 mole of magnesium nitride.
Therefore no. of moles of nitrogen = no. of moles of magnesium nitride.
Molar mass of magnesium nitride = 101gmol^-1
the no. of moles = 2moles
\[ mass = moles* molar mass\]
\[mass = 2* 101 = 202g\]

Answer 5

thank you !