Question

How many grams of aluminum sulfite will be formed from 125mL of a 0.25 M solution of sulfurous acid. Complete this question by ansering the following:

a ) How many moles of sulfurous acid are present in 125 mL of a 0.25M solution?

b) What is the molar ratio of sulfurous acid to aluminum sulfite?

c) How many moles of aluminum sulfite will be produced?

d) What is the molar mass of aluminum sulfite?

e) How many frams of aluminum sulfite will be produced?

Answer 1

2Al + 3H2SO3 --------> Al2(So3)3 + 3H2

a. no of moles of H2SO3 = molarity * volume in L

                                         = 0.25*0.125 = 0.03125 moles

b. 2Al + 3H2SO3 --------> Al2(So3)3 + 3H2

             3 moles               1 mole

           3*82                      294

     H2So3 : Al2(SO3)3   = 3*82 : 294

                                       = 0.836 :1

c. 2Al + 3H2SO3 --------> Al2(So3)3 + 3H2

   3 moles of H2So3 react with AL to gives 1 mole of Al2(SO3)3

   0.03125 moles of H2So3 react with Al to gives =1*0.03125/3 = 0.0104 moles of Al2(SO3)3

d. molar mass of Al2(sO3)3 = 27*2 + 32*3 + 16*9 = 294g/mole

e. mass of Al2(SO3)3 =no of moles * gram molar mass

                                     = 0.0104*294 = 3.0576g