In the combustion reaction of methane gas : CH4 (g) + O2 (g) --> CO2 (g) + 2 H2O (g) 4.0 X 10 exp 23 molecules of CH4 was used. Determine

(a) the moles of CO2 produced

(b) using your answer in (a), calculate the volume of CO2 produced at STP

(c) the volume of oxygen used in the reaction at STP

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explanation:

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole