Application 2: we mixed 400 g of carbon dioxide CO2 with water H2O, oxygen and glucose C6H12O6 are formed according to the following chemical equation: CO2 (g) + H2O C6H12O6 + O2 (g)
1- Balance the above reaction.
2- Calculate the number of mole of water H2O needed, and that of products.
3- Deduce the mass of C6H12O6 and the volume of oxygen O2 released.
Given: molar mass in g. mol-1: C: 12 H: 1 O: 16

h let me think

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