Question

If 20.0g of CO2 and 4.4g of CO2

areplaced in a 5.00 L container at 21oC, what is
thepressure of this mixture of gases?

Answer 1

The given question is incorrect. The correct question is as follows.

If 20.0 g of
`O_(2)` and 4.4 g of
`CO_(2)` are placed in a 5.00 L container at
`21^(o)C`, what is the pressure of this mixture of gases?

Step-by-step explanation:

As we know that number of moles equal to the mass of substance divided by its molar mass.

Mathematically, No. of moles =
`\frac{\text{mass}}{\text{molar mass}}`

Hence, we will calculate the moles of oxygen as follows.

No. of moles =
`\frac{\text{mass}}{\text{molar mass}}`

Moles of
`O_(2)` =
`(20.0 g)/(32 g/mol)`

= 0.625 moles

Now, moles of
`CO_(2) = (4.4 g)/(44 g/mol)`

= 0.1 moles

Therefore, total number of moles present are as follows.

Total moles = moles of
`O_(2)` + moles of
`CO_(2)`

= 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 =
`(nRT)/(V)`

=
`(0.725 mol * 0.08206 Latm/mol K * 294 K)/(5.00 L)`

=
`(17.491089)/(5)` 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.