Answer:
the volume occupied by 3.0 g of the gas is 16.8 L.
Explanation:
Given;
initial reacting mass of the helium gas, m₁ = 4.0 g
volume occupied by the helium gas, V = 22.4 L
pressure of the gas, P = 1 .0 atm
temperature of the gas, T = 0⁰C = 273 K
atomic mass of helium gas, M = 4.0 g/mol
initial number of moles of the gas is calculated as follows;
The number of moles of the gas when the reacting mass is 3.0 g;
m₂ = 3.0 g
The volume of the gas at 0.75 mol is determined using ideal gas law;
PV = nRT
Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.