For the following reaction, 3.70 grams of oxygen gas are mixed with excess carbon monoxide . the reaction yields 8.25 grams of carbon dioxide . carbon monoxide(g) + oxygen(g) carbon dioxide(g) what is the ideal yield of carbon dioxide? grams what is the percent yield for this reaction? %

81% Yield

Explanation:

2CO        +               O₂                                =>     2CO₂

Excess                3.70g O₂                          =>     8.25g CO₂ (actural yield)

               (3.70g O₂)/(32g O₂/mol O₂)

                   = 0.1156 mol O₂                       =>  2(0.1156) mol CO₂

                                                                          = 10.175g (Theoretical Yield)

%Yield = (Actual Yield / Theoretical Yield)100%  

= (8.25g/10.175g)100% = 81% Yield

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