Question

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

Answer 1

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(O₂) = 34.72 L.

Answer 2

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