Question

Question #73c3c

Answer 1

23,5 L

Explanation:

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`

Answer 2

The volume of nitrogen is C) `"27.7 dm"^3`

Explanation:

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.