Question

When 0.500 mole of sodium is produced according to the reaction `2NaN_3(s) -> 3N_2(g) + 2Na(s)`, how many moles of nitrogen gas are also produced?

Answer 1

`0.750mol` of nitrogen gas is produced.

Explanation:

You can solve quite easily by putting in an `n`.

`2nNaN_3 -> 3nN_2 + 2nNa`

We know that 0.500 moles of `Na` are produced. We also know that `2n` amounts of `Na` are produced, so

`0.500 mol = 2n`

`n = 0.250 mol`

Now we know that there are `3nN_2` produced, so

`3n = 3 * 0.250mol = 0.750mol`