# 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?

Mar 1, 2017

#### Answer:

$0.750 m o l$ of nitrogen gas is produced.

#### Explanation:

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

$2 n N a {N}_{3} \to 3 n {N}_{2} + 2 n N a$

We know that 0.500 moles of $N a$ are produced. We also know that $2 n$ amounts of $N a$ are produced, so

$0.500 m o l = 2 n$

$n = 0.250 m o l$

Now we know that there are $3 n {N}_{2}$ produced, so

$3 n = 3 \cdot 0.250 m o l = 0.750 m o l$