Question

For this reaction, 25.0 g sulfur dioxide reacts with 5.91 g water. sulfur dioxide (g) + water (l) sulfurous acid (H2SO3) (g) What is the maximum mass of sulfurous acid (H2SO3) that can be formed? g What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete?

Answer 1

SO2 + H2O --------------------------> H2SO3

25g/64g/mol=0.39 5.91g/18g/mol=0.328 0 initial moles

0.39-0.328=0.0617 0 0.328 equilibrium

As the moles  of H2O are less than SO2 ( both reacting in 1:1ratio) H2O is the limiting reagent .

the excess reagent is SO2

amount of excess reagent remaining = moles remaining x molar mass

= 0.0617 molx 64g/mol

= 3.947 g