For the reaction

6Li + N2 - > 2Li3N

What us the maximum amount of Li3B which could be formed from 7.87 mol of Li and 4.37 mol of N2?

Answer in units of mol.

Mass = 90.6 g

2.6 mol

Explanation:

Given dа ta:

Number of moles of Li = 7.87 mol

Number of moles of N = 4.37 mol

Mass of Li₃N = ?

Solution

Chemical equation:

6Li + N₂ → 2Li₃N

Now we will compare the moles of nitrogen and lithium with Li₃N.

N₂ : Li₃N

1 : 2

4.37 : 2*4.37 = 8.74

Li : Li₃N

6 : 2

7.87 : 2/6*7.87 = 2.6 mol

Mass of Li₃N:

Mass = number of moles * molar mass

Mass = 2.6 mol * 34.83 g/mol

Mass = 90.6 g