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

We need 51.7 grams of NaN3

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

c is the answer because when something is left after something evaporates that shows the chemical change.