What volume does 43.5 g of n2 occupy at stp? (r = 0.08206 l⋅atm/mol⋅k) ?

Answer is: volume of nitrogen is 34.72 liters.

m (N₂) = 43.5 g.

n (N₂) = m (N₂) : M (N₂).

n (N₂) = 43.5 g : 28 g/mol.

n (N₂) = 1.55 mol, amount of substance.

T = 273 K, standard temperature.

p = 1 atm, standard pressure.

R = 0.08206 L·atm/mol·K, universal gas constant.

Ideal gas law: p·V = n·R·T.

V = n·R·T / p.

V (N₂) = 1.55 mol · 0.08206 L·atm/mol·K · 273 K / 1.00 atm.

V (O ₂) = 34.72 L.

We are going to use this formula:

PV = n RT

when at STP

p is the pressure = 1 atm

and n is the moles = mass / molar mass = 43.5 g / 28g/mol

= 1.56 moles

R is the ideal gas constant = 0.0821

and T is a temperature in Kelvin = 273 K

by substitution, we will get V (the volume)

1 atm * V = 1.56 moles * 0.0821 * 273K

∴ V = 34.96 L