If 95 grams of Iron are needed to react with sulfur, how many grams of sulfur are involved in this reaction? 2 Fe + 3 S à Fe2S3

You need a balanced equation.

2Fe(s) + 3S(s) ==> Fe2S3(s) ... balanced equation

Reactants are solid iron and solid sulfur

Product is iron(III) sulfide

A. 1.42 moles Fe x 3 moles S/2 moles Fe = 2.13 moles sulfur

B. 2.75 moles S x 2 moles Fe/3 moles S = 1.83 moles iron

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