Answer to Question #106589 in Chemistry for randall

Question

How many grams of sodium azide are required to produce 28.0 g of nitrogen?

Answer 1

Solution:

The decomposition of sodium azide (NaN₃) is:

2NaN₃ => 2Na(s) + 3N₂(g)

According to the chemical equation: n(NaN₃)/2 = n(Na)/2 = n(N₂)/3.

n(NaN₃)/2 = n(N₂)/3;

3*n(NaN₃) = 2*n(N₂).

a)

m(NaN₃) = ??? g;

M(NaN₃) = Ar(Na) + 3*Ar(N) = 23 + 3*14 = 65 (g/mol);

n(NaN₃) = m(NaN₃) / M(NaN₃);

n(NaN₃) = m(NaN₃) / 65

b)

m(N₂) = 28.0 g;

M(N₂) = 2*Ar(N) = 2*14 = 28 (g/mol);

n(N₂) = m(N₂) / M(N₂);

n(N₂) = 28.0 / 28 = 1.0 mol

Consequently,

3*n(NaN₃) = 2*n(N₂);

n(NaN₃) = 2*n(N₂) / 3

m(NaN₃) / 65 = (2 * 1.0) / 3

m(NaN₃) / 65 = 0.667;

m(NaN₃) = 65*0.667 = 43.35

m(NaN₃) = 43.35 g

Answer: 43.35 g of sodium azide are required to produce 28.0 g of nitrogen.