Question

1)

For the following reaction. 25.6 grams of sulfur dioxide are allowed to react with 6.09 grams of water. sulfur dioxide (g) + water (1) sulfurous acid (H2SO3) (9) 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

2)

I don't need detailed explanation. I just need answers, please make sure they are correct and written in the same way they need them up there Thank You!!!

Answer 1

1. Sol :-

Given, mass of SO₂ = 25.6 g

Gram molar mass of SO₂ = 64 g/mol

Number of moles of SO₂ gas = Given mass of SO₂ / Gram molar mass of SO₂

= 25.6 g / 64 g/mol

= 0.40 mol

Similarly,

mass of H₂O = 6.09 g

Gram molar mass of H₂O = 18 g/mol

Number of moles of H₂O = Given mass of H₂O / Gram molar mass of H₂O

= 6.09 g / 18 g/mol

= 0.338 mol

ICF table of the given reaction is :

..............................SO₂ (g).............+............H₂O (l) <--------------------> H₂SO₃ (g)

Initial (I)..................0.40 mol..........................0.338 mol..........................0.0 M

Change (C)............-0.338 mol.......................-0.338 mol..........................+0.338 mol

Final (F)...................0.062 mol.........................0.0 mol.................................0.338 mol

Limiting reagent formula = H2O

Excessive reagent = SO₂ (g) Limiting reagent :- Which is completely consumed in a chemical reaction.

Excessive reagent = Whose amount remains left after the completion of chemical reaction

Mass of excessive reagent left = Moles x Gram molar mass

= 0.062 mol x 64 g/mol

= 3.968 g

Hence, Mass of excessive reagent remains in the solution = 3.968 g

Mass of products formed = Moles x Gram molar mass

= 0.338 mol x 82.07 g/mol

= 27.740 g

Hence, Maximum amount of H2SO3 formed = 27.740 g