Determine the limiting reactant and calculate the number of grams of sulfur dioxide, SO,, that can be formed when 27.3 g of methane thiol, CH, SH, reacts with 38.6 g of oxygen, O2. The balanced chemical equation and
molar masses are given below.
CH, SH+3 02600 - 2 H,06+ CO260 + SO218)

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

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