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?

1 Answer
Mar 1, 2017

Answer:

#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#