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?

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Answer 1 · 2017-03-01T16:12:28

$0.750mol$ of nitrogen gas is produced.

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

$3n = 3 * 0.250mol = 0.750mol$

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