Question #f05aa

1 Answer
Sep 16, 2016

"18.6 g NaN"_3" are required to form "12.0 g N"_2"

Explanation:

Balanced Equation

"2NaN"_3("s")rarr"2Na(s)"+"3N"_2("g")

We will be doing the following pattern to answer this question.

"mass N"_2"rarr"mol N"_2"rarr"mol NaN"_3"rarr"mass NaN"_3"

We need the molar ratios between "NaN"_3 and "N"_2.

From the equation, we can see that there are two mole ratios:

(2"mol NaN"_3)/(3"mol N"_2 ") and (3"mol N"_2)/(2"molNaN"_3)

We need to determine the molar masses of the nitrogen gas and sodium azide.

Molar Mass "N"_2":(2xx"14.0067 g/mol)="28.0134 "g/mol N"_2"

Molar Mass "NaN"_3:""65.009869 g/mol NaN"_3"
https://www.ncbi.nlm.nih.gov/pccompound?term=NaN3

Next we need to determine how many moles are in "12.0 g N"_2" by dividing Its given mass by its molar mass. We will need to do the same for "NaN"_3".

Moles "N"_2":12.0 cancel"g N"_2xx"1 mol N"_2/(28.0134 cancel"g N"_2)="0.4284 mol N"_2"

To determine the moles of "NaN"_3", we need to multiply the mole ratio from the equation. We will need to use the mole ratio from the equation with "NaN"_2" in the numerator.

0.4284 cancel"mol N"_2xx(2"mol NaN"_3)/(3cancel"mol N"_2)="0.2856 mol NaN"_3"

Now we need to multiply the moles "NaN"_3" by its molar mass.

0.2856 cancel"mol NaN"_3""xx(65.009869"g NaN"_3)/(1cancel"mol NaN"_3)="18.6 g NaN"_3" (rounded to three significant figures.