In this reaction: Mg (s) + I₂ (s) → MgI₂ (s), if 10.0 g of Mg reacts with 60.0 g of I₂, and 53.88 g of MgI₂ form, what is the percent yield?

We know the law of conservation of mass

Here

Lets find the percentage

може ли да зареждате повече информация, моля

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