Limiting Reactants: Compare Product Method (Mass/Mass)

Question

For the following reaction, 28.1 grams of sulfur dioxide are allowed to react with 6.69 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

Hi Omar,

The formula for sulfur dioxide is SO₂. Sulfur dioxide's molar mass is 32.065 g/mol + 2(15.999 g/mol) = 64.063 g/mol.

The formula for water is H₂O. Water's molar mass is 2(1.008 g/mol) + 15.999 g/mol = 18.015 g/mol.

This means that there are 28.1 g / 64.063 g/mol = 0.4386 mol of sulfur dioxide to start, and there are 6.69 g / 18.015 g/mol = 0.3714 mol of water to start. For every molecule of sulfur dioxide that is consumed, one water molecule will be consumed to form one sulfurous acid molecule. Because there are fewer moles of water to start, the water (H₂O) will be the limiting reagent. The maximum amount of sulfurous acid that can be formed depends on the amount of water in moles to start. 0.3714 mol of sulfurous acid can be formed, and the molar mass of sulfurous acid is 18.015 g/mol + 64.063 g/mol = 82.078 g/mol, so the maximum amount of sulfurous acid that can be formed is 82.078 g/mol * 0.3714 mol = 30.5 g (rounded to 3 significant figures).

The amount of excess reagent is equal to 0.4386 mol - 0.3714 mol = 0.0672 mol SO₂. In grams, this is 0.0672 mol * 64.063 g/mol = 4.31 g.

Hope that helps!