Balanced Equation
"2NaN"_3("s")2NaN3(s)rarr→"2Na(s)"+"3N"_2("g")2Na(s)+3N2(g)
We will be doing the following pattern to answer this question.
"mass N"_2"mass N2rarr→"mol N"_2"mol N2rarr→"mol NaN"_3"mol NaN3rarr→"mass NaN"_3"mass NaN3
We need the molar ratios between "NaN"_3NaN3 and "N"_2N2.
From the equation, we can see that there are two mole ratios:
(2"mol NaN"_3)/(3"mol N"_2 ")2mol NaN33mol N2 and (3"mol N"_2)/(2"molNaN"_3)3mol N22molNaN3
We need to determine the molar masses of the nitrogen gas and sodium azide.
Molar Mass "N"_2":N2:(2xx"14.0067 g/mol)="28.0134 "g/mol N"_2"(2×14.0067 g/mol)=28.0134g/mol N2
Molar Mass "NaN"_3:"NaN3:"65.009869 g/mol NaN"_3"65.009869 g/mol NaN3
https://www.ncbi.nlm.nih.gov/pccompound?term=NaN3
Next we need to determine how many moles are in "12.0 g N"_2"12.0 g N2 by dividing Its given mass by its molar mass. We will need to do the same for "NaN"_3"NaN3.
Moles "N"_2":N2:12.0 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.
