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.

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

$\text{Moles of }CO_2=\frac{12g}{44g/mol}=0.273mol$

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:

$P\times (10.0L)=(0.273mol)\times (0.0821L.atm/mol.K)\times (265K)$

$P=0.59atm$

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