Question

A reaction between liquid reactants takes place at -8.0 C in a sealed, evacuated vessel with a measured volume of 10.0 L. Measurements show that the reaction produced 12 g or carbon dioxide gas. Calculate the pressure of carbon dioxide in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round you answer to 2 significant digits.

Answer 1

Answer : The pressure of carbon dioxide in the reaction vessel after the reaction is, 0.59 atm.

Explanation : Given,

Mass of
`CO_2` = 12 g

Molar mass of
`CO_2` = 44 g/mol

First we have to calculate the moles of carbon dioxide gas.

`\text{Moles of }CO_2=\frac{\text{Given mass of }CO_2}{\text{Molar mass of }CO_2}`

`\text{Moles of }CO_2=(12g)/(44g/mol)=0.273mol`

Now we have to calculate the pressure of carbon dioxide gas.

Using ideal gas equation:

`PV=nRT`

where,

P = pressure of gas = ?

V = volume of gas = 10.0 L

n = number moles of gas = 0.273 mol

R = gas constant = 0.0821 L.atm/mol.K

T = temperature of gas =
`-8.0^oC=273+(-8.0)=265K`

Now put all the given values in the above formula, we get:

`P* (10.0L)=(0.273mol)* (0.0821L.atm/mol.K)* (265K)`

`P=0.59atm`

Therefore, the pressure of carbon dioxide in the reaction vessel after the reaction is, 0.59 atm.