What volume of nitrogen gas is produced by the explosion of 65.0 g of sodium azide at STP? The balanced chemical equation is 2 NaN3 (s) = 2 Na (s) + 3 N2 (g) for this reaction.

a. 33.6 L

b. 67.2 L

c. No right choice.

a) V = 33.6 L

Explanation:

Given dа ta:

Mass of sodium azide = 65 g

Volume of Nitrogen = ?

Solution:

Chemical equation:

2NaN₃ → 3N₂ + 2Na

Number of moles of NaN₃:

Number of moles of NaN₃ = Mass / molar mass

Number of moles of NaN₃ = 65 g / 65 g/mol

Number of moles of NaN₃ = 1 mol

Now we will compare the moles of NaN₃ with N₂.

NaN₃ : N₂

2 : 3

1 : 3/2 = 1.5 mol

Volume of nitrogen gas:

PV = nRT

V = nRT / P

V = 1.5 mol. 0.0821 atm. L. mol⁻¹. K⁻¹ * 273.15 K / 1 atm

V = 33.64 L