Question #73c3c

50 g of sodium azide #NaN_3# (MM= 71g/mol) , correspond to #(50g)/(71 g/(mol))= 0,70 mol.#,

From the balanced reaction you can see that from 2 mol of #NaN_3# you obtain 3 mole of #N_2# so, making the proportion, from 0,70 mol you can obtain 1,05 mol of nitrogen.

As 1 mol of any gas occupies a volume of 22,4 L in normal condition (P= 1 atm and T= 273K), 1,05 mol occupy #22,4 L/(mol) xx 1,05 mol = 23,5 L#

The balanced chemical equation is

#"2NaN"_3 → "3N"_2 + "2Na"#

Step 1. Calculate the moles of #"NaN"_3#

#"Moles of NaN"_3 = 50 color(red)(cancel(color(black)("g NaN"_3))) × "1 mol NaN"_3/(65.01 color(red)(cancel(color(black)("g NaN"_3)))) = "0.769 mol NaN"_3#

Step 2. Calculate the moles of #"N"_2#

#"Moles of N"_2 = 0.769 color(red)(cancel(color(black)("mol NaN"_3))) × "3 mol N"_2/(2 color(red)(cancel(color(black)("mol NaN"_3)))) = "1.15 mol N"_2#

Step 3. Calculate the volume of #"N"_2#

We aren't given the volume or the temperature, so we will have to make an assumption.

Let's assume NTP (1 atm and 20 °C).

Then we can use the Ideal Gas Law to calculate the volume:

#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#

We can rearrange the Ideal Gas Law to get

#V = (nRT)/p#

In this problem,

#n = "1.15 mol"#
#R = "0.082 06 dm"^3·"atm·K"^"-1""mol"^"-1"#
#T = "(20 + 273.15) K = 293.15 K"#
#p = "1 atm"#

#V = (1.15 color(red)(cancel(color(black)("mol"))) × "0.082 06 dm"^3·color(red)(cancel(color(black)("atm"·"K"^"-1"·"mol"^"-1"))) × 293.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("atm")))) = "27.7 dm"^3#

Note: The answer should have only two significant figures, because that is all you gave for the mass of sodium azide.

However, I calculated to three significant figures, because that is the answer in
Option C.