A reaction between liquid reactants takes place at in a sealed, evacuated vessel with a measured volume of . Measurements show that the reaction produced of carbon monoxide gas. Calculate the pressure of carbon monoxide gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Be sure your answer has the correct number of significant digits.

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Complete Question

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Answer:

The pressure is  [tex]P = 0.76 \ atm[/tex]

Explanation:

From the question we are told that

   The mass of the carbon monoxide  is  [tex]m_c = 17 \ g[/tex]

   The temperature at which takes place  [tex]T = 30 .0^oC = 30 + 273 = 303 K[/tex]

   The volume of the sealed vessel  is  [tex]V = 20 .0L[/tex]

Generally the ideal gas law is mathematically represented as

        [tex]PV = nRT[/tex]

Where R is the gas constant with value  [tex]R = 0.0821 \ L \ atm \ mol^{-1} K^{-1}[/tex]

n is the number of moles of carbon monoxide  which is mathematically evaluated as

             [tex]n = \frac{m_c}{M_c}[/tex]

where [tex]M_c[/tex] is the molar mass of carbon monoxide which is a constant with value

        [tex]M_c = 28 \ g/mol[/tex]

So     [tex]n = \frac{17}{28}[/tex]

         [tex]n =0.6071[/tex]

Now Making P the subject we have

           [tex]P =\frac{ 0.6071 * 0.0821 * 303}{20}[/tex]

           [tex]P = 0.76 \ atm[/tex]

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