# Question #73c3c

##### 2 Answers

#### Answer:

23,5 L

#### Explanation:

50 g of sodium azide

From the balanced reaction you can see that from 2 mol of

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

#### Answer:

The volume of nitrogen is C)

#### Explanation:

The balanced chemical equation is

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

**Step 2. Calculate the moles of #"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,

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.