Answer:
b. 15.39 L
Step-by-step explanation:
Given;
reacting mass of the Nitrogen gas Nā, m = 42 g
pressure of the gas, P = 2 atm
temperature of the gas, T = 250 K
molar mass of Nitrogen, N = 14 g/mol
molar mass of Nitrogen gas Nā, M = 28 g/mol
number of moles of the given Nitrogen gas;
[tex]n = \frac{Reacting \ mass}{Molar \ mass} = \frac{m}{M} \\\\n = \frac{42 \ g}{28 \ g/mol} = 1.5 \ mol.[/tex]
The volume occupied by the Nitrogen gas can be calculated from Ideal gas law;
PV = nRT
where;
V is the volume of the gas
R is ideal gas constant = 0.0821 atm.L / mol.K
From the equation above, make V the subject of the formula;
[tex]V = \frac{nRT}{P} \\\\V = \frac{1.5\ mol. \ \times \ 0.0821 \ atm.L/mol.K \ \times \ 250 \ K}{2 \ atm} \\\\V = 15.39 \ L[/tex]
Therefore, the volume occupied by the Nitrogen gas is 15.39 L
