Question

If 80.0g of sodium is mixed with 40.0g of nitrogen gas, what mass of sodium nitride forms?

Answer 1

You need to put this on the science subject

Answer 2

but anyways,

This is a limiting reactant problem, so the first step is to determine which substance is the limiting reactant, and then to use that to determine how much sodium nitride is made.

Molar mass Na = 23.0 g/mol
Molar mass N2 = 14.0 g/mol * 2 = 28.0 g/mol <-- doubled because of the two nitrogens
Molar mass Na3N = 23.0 * 3 + 14.0 = 83.0 g/mol

80.0 g Na * (1 mol / 23.0 g) = 3.48 mol Na * (2 mol Na3N / 6 mol Na) = 1.16 mol Na3N
40.0 g N2 * (1 mol / 28.0 g) = 1.43 mol N2 * (2 mol Na3N / 1 mol N2) = 2.86 mol Na3N
Therefore, sodium is the limiting reactant, so it will be used to determine how much Na3N is made.
1.16 mol Na3N * (83.0 g / mol) = 96.2 g Na3N
If the reaction only yields 85%, just multiply the answer by .85, to get 81.8g Na3N.

YW