Question

If 20.0g of CO2 and 4.4g of CO2

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?

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 $O_{2}$ = $\frac{20.0 g}{32 g/mol}$

                            = 0.625 moles

Now,   moles of $CO_{2} = \frac{4.4 g}{44 g/mol}$

                                      = 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 = $\frac{nRT}{V}$

                   = $\frac{0.725 mol \times 0.08206 Latm/mol K \times 294 K}{5.00 L}$

                    = $\frac{17.491089}{5}$ 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.