Question #73c3c

2 Answers
Jul 14, 2017

23,5 L

Explanation:

50 g of sodium azide NaN_3NaN3 (MM= 71g/mol) , correspond to (50g)/(71 g/(mol))= 0,70 mol.50g71gmol=0,70mol.,

From the balanced reaction you can see that from 2 mol of NaN_3NaN3 you obtain 3 mole of N_2N2 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 L22,4Lmol×1,05mol=23,5L

Jul 15, 2017

The volume of nitrogen is C) "27.7 dm"^327.7 dm3

Explanation:

The balanced chemical equation is

"2NaN"_3 → "3N"_2 + "2Na"2NaN33N2+2Na

Step 1. Calculate the moles of "NaN"_3NaN3

"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.