If 2.18 grams of Br2 react with excess KI, how many grams of KBr are formed? Br2 + 2 K1 - 12 +2 KB

2.18 grams of Br2 reacts to produce 3.25 grams of KBr

Explanation:

The balanced equation is

2 KI + Br2 → 2 KBr + I2

2 moles of KI and 1 mole of Br2 reacts to produce two molecules of KBr

Mass of two moles of KI = 2 * 166 grams

Mass of one mole of Br2 = 159.808 grams

Mass of two moles of KBr = 2* 119 grams

Thus,  159.808 grams of Br2 reacts to produce 2* 119 grams of KBr

2.18 grams of Br2 reacts to produce {(2* 119)/ 159.808} * 2.18 grams of KBr

2.18 grams of Br2 reacts to produce 3.25 grams of KBr

just did the assignment and the answer was #1 +2 n

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