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.

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Alleei Alleei · Answer 1 · 2020-04-24T04:31:12+02:00

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

Explanation : Given,

Mass of [tex]CO_2[/tex] = 12 g

Molar mass of [tex]CO_2[/tex] = 44 g/mol

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

[tex]\text{Moles of }CO_2=\frac{\text{Given mass of }CO_2}{\text{Molar mass of }CO_2}[/tex]

[tex]\text{Moles of }CO_2=\frac{12g}{44g/mol}=0.273mol[/tex]

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

Using ideal gas equation:

[tex]PV=nRT[/tex]

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 = [tex]-8.0^oC=273+(-8.0)=265K[/tex]

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

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

[tex]P=0.59atm[/tex]

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