2 Al + 3 FeO → 3 Fe + Al₂O₃ If 51.0 g of Al was mixed with excess FeO, how much Fe could be theoretically collected? 1. 70.3 g Fe
2. 105 g Fe
3. 158.3 g Fe
4. 2,848 g Fe

Al + 3 FeO → 3 Fe + Al₂O₃ Using the same information as above, the ACTUAL YIELD of the Fe was 125.7 g. What is the percent yield?

1. 4.41%
2. 55.9%
3. 79.4%
4. 83.5%

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

4d^55s^1 ( third choice)