Question

For the following reaction, 22.6 grams of sulfur dioxide are allowed to react with 10.6 grams of water.

sulfur dioxide (g) + water (l) ---> sulfurous acid (H₂SO₃) (g)

What is the maximum amount of sulfurous acid (H₂SO₃) that can be formed?

grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

     SO2 + H2O --------------> H2SO3

no of moles of SO2 = W/G.M.Wt

                                = 22.6/64   = 0.353 moles

no of moles of H2O   = W/G.M.Wt

                                   = 10.6/18   = 0.59 moles

from balanced equation

1 moles of SO2 react with 1 moles H2O

0.353 moles of SO2 react with 0.353 moles of H2O

H2O is excess reagent

the excess reagent remains after the reaction is complete   = 0.59-0.353   = 0.237moles

amount of the excess reagent remains after the reaction is complete = no of moles * gram molar mass

                                                                                                             = 0.237*18   = 4.27 g of H2O

limiting reagent is SO2

1 moles of SO2 react with H2O to gives 1moles H2SO3

0.353 moles of SO2 react with H2O to gives 0.353 moles of H2SO3

mass of H2So3   = no of moles * gram molar mass

                            = 0.353*82   = 28.95g of H2So3