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.
