5g sodium reacts with nitrogen. find m and n product

Given:
m (Na) = 5 g

Find:
n (Na3N) -?
m (Na3N) -?

Solution:
1) 6Na + N2 => 2Na3N;
2) M (Na) = Mr (Na) = Ar (Na) = 23 g / mol;
M (Na3N) = Mr (Na3N) = Ar (Na) * 3 + Ar (N) = 23 * 3 + 14 = 83 g / mol;
3) n (Na) = m (Na) / M (Na) = 5/23 = 0.22 mol;
4) n (Na3N) = n (Na) * 2/6 = 0.2 * 2/6 = 0.07 mol;
5) m (Na3N) = n (Na3N) * M (Na3N) = 0.07 * 83 = 5.81 g.

Answer: The amount of Na3N substance is 0.07 mol; weight – 5.81 g.