Calculate the mass of the reaction product formed by the interaction of sodium with nitrogen with a volume of 6.72 liters.

The product of the interaction of these substances is sodium azide:

2Na + 3N2 = 2NaN3 (do not forget to place the coefficients for subsequent correct calculations);

Through the volume of nitrogen given to us, we can find the amount of its mol (n):

n (N2) = V: Vm (molar volume) = 6.72 l: 22.4 l / mol = 0.3 mol (22.4 l is a constant value and the same for all gases under normal conditions).

Sodium azide is 1.5 times less than nitrogen (3: 2 = 1.5, based on the coefficients), therefore n (NaN3) = 0.3 mol: 1.5 = 0.2 mol.

It remains to find the mass of NaN3:

m (NaN3) = n (NaN3) * M (NaN3) = 0.2 mol * 65 g / mol = 13 g.