Question

Solid sodium azide (NaN3) produces solid sodium and nitrogen gas. How many grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres

Answer 1

We need 51.7 grams of NaN3

Step-by-step explanation:

Step 1 : Data given

Volume of N2 gas produced = 26.5 L

Temperature = 295 K

Pressure = 1.10 atm

Step 2: The balanced equation

2NaN3 → 2Na + 3N2

Step 3: Calculate moles N2

p*V = n*R*T

⇒with p = the pressure of nitrogen gas = 1.10 atm

⇒with V = volume of nitrogen gas = 26.5 L

⇒with n = the number of moles nitrogen gas = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol* K

⇒with T = the themperature = 295

n = (p*V)/(R*T)

n = (1.10 * 26.5) / (0.08206 * 298)

n = 1.192 moles N2

Step 4: Calculate moles NaN3

For 2 moles NaN3 we'll have 2 moles Na and 3 moles N2

For 1.192 moles N2 we need 2/3 * 1.192 = 0.795 moles NaN3

Step 5: calculate mass NaN3

Mass NaN3 = moles NaN3 * molar mass NaN3

Mass NaN3 = 0.795 moles * 65.0 g/mol

Mass NaN3 = 51.7 gtams NaN3

We need 51.7 grams of NaN3

Answer 2

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

Step-by-step explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P*V = n*R*T

In this case, the balanced reaction is:

2 NaN₃ → 2 Na + 3 N₂

You know the following about N₂:

• P= 1.10 atm
• V= 26.5 L
• n=?

• 
  `R=0.082057 (atm*L)/(mol*K)`

• T= 295 K

Replacing in the equation for ideal gas:

1.10 atm* 26.5 L= n*
`0.082057 (atm*L)/(mol*K)`*295 K

Solving:

`n=(1.10 atm*26.5 L)/(0.082057 (atm*L)/(mol*K) *295K)`

n= 1.2 moles

Now, the following rule of three can be applied: if 3 moles of N₂ are produced by stoichiometry of the reaction from 2 moles of NaN₃, 1.2 moles of N₂ are produced from how many moles of NaN₃?

`moles of NaN_(3)=(1.2 molesofN_(2) *2 molesofNaN_(3) )/(3 molesofN_(2) )`

moles of NaN₃= 0.8

Since the molar mass of sodium azide is 65.01 g / mol, then one last rule of three applies: if 1 mol has 65.01 grams of NaN₃, 0.8 mol how much mass does it have?

`mass of NaN_(3) =(0.8 mol*65.01 grams)/(1 mol)`

mass of NaN₃=52.008 grams

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.