How many grams of sodium azide are needed to provide sufficient nitrogen gas to fill a 50.0 × 50.0 × 25.0 cm bag to a pressure of 1.15 atm at 25.0 °C? 

1 Answer
Jul 29, 2017

Answer:

#127# #"g NaN"_3#

Explanation:

We're asked to find the mass, in #"g"#, of sodium azide (#"NaN"_3#) needed to produce a certain amount of #"N"_2#.

To do this, we can use the ideal gas equation to find the moles of nitrogen gas present:

#PV = nRT#

  • #P = 1.15# #"atm"# (given)

  • #V# must be in liters, so we can find the volume first in #"cm"^3# and then convert:

#50.0 xx 50.0 xx 25.0 = 62500# #"cm"^3#

#62500cancel("cm"^3)((1cancel("mL"))/(1cancel("cm"^3)))((1color(white)(l)"L")/(10^3cancel("mL"))) = 62.5# #"L"#

  • #R# is the universal gas constant, equal to #0.082057("L"·"atm")/("mol"·"K")#

  • #T = 25.0# #""^"o""C"#, which must be in Kelvin:

#T = 25.0# #""^"o""C" + 273 = 298# #"K"#

Plugging in known values, and solving for the number of moles, #n#, we have

#n = (PV)/(RT) = ((1.15cancel("atm"))(62.5cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(298cancel("K"))) = color(red)(2.94# #color(red)("mol N"_2#

Now, we can use the coefficients of the chemical equation to find the relative number of moles of sodium azide that must react:

#color(red)(2.94)cancel(color(red)("mol N"_2))((2color(white)(l)"mol NaN"_3)/(3cancel("mol N"_2))) = color(green)(1.96# #color(green)("mol NaN"_3#

Finally, we can use the molar mass of sodium azide (#65.01# #"g/mol"#) to find the mass in grams:

#color(green)(1.96)cancel(color(green)("mol NaN"_3))((65.01color(white)(l)"g NaN"_3)/(1cancel("mol NaN"_3))) = color(blue)(ul(127color(white)(l)"g NaN"_3#