Respuesta :
The given question is incorrect. The correct question is as follows.
If 20.0 g of [tex]O_{2}[/tex] and 4.4 g of [tex]CO_{2}[/tex] are placed in a 5.00 L container at [tex]21^{o}C[/tex], what is the pressure of this mixture of gases?
Explanation:
As we know that number of moles equal to the mass of substance divided by its molar mass.
Mathematically, No. of moles = [tex]\frac{\text{mass}}{\text{molar mass}}[/tex]
Hence, we will calculate the moles of oxygen as follows.
No. of moles = [tex]\frac{\text{mass}}{\text{molar mass}}[/tex]
Moles of [tex]O_{2}[/tex] = [tex]\frac{20.0 g}{32 g/mol}[/tex]
= 0.625 moles
Now, moles of [tex]CO_{2} = \frac{4.4 g}{44 g/mol}[/tex]
= 0.1 moles
Therefore, total number of moles present are as follows.
Total moles = moles of [tex]O_{2}[/tex] + moles of [tex]CO_{2}[/tex]
= 0.625 + 0.1
= 0.725 moles
And, total temperature will be:
T = (21 + 273) K = 294 K
According to ideal gas equation,
PV = nRT
Now, putting the given values into the above formula as follows.
P = [tex]\frac{nRT}{V}[/tex]
= [tex]\frac{0.725 mol \times 0.08206 Latm/mol K \times 294 K}{5.00 L}[/tex]
= [tex]\frac{17.491089}{5}[/tex] atm
= 3.498 atm
or, = 3.50 atm (approx)
Therefore, we can conclude that the pressure of this mixture of gases is 3.50 atm.
