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

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).