Question

Assuming you react 38.3 g of sulfur dioxide with 32.8 g of sodium hydroxide and assuming that the reaction goes to completion, calculate the mass of sodium sulfite that formed.

Answer 1

Answer:

51.7 g

Explanation:

The reaction that takes place is:

• SO₂ + 2NaOH → Na₂SO₃ + H₂O

First we convert the mass of each reactant into moles, using their respective molar mass:

• SO₂ ⇒ 38.3 g ÷ 64.066 g/mol = 0.598 mol

• NaOH ⇒ 32.8 g ÷ 40 g/mol = 0.82 mol

0.598 mol of SO₂ would react completely with (2*0.598) 1.196 moles of NaOH. There are not as many NaOH moles, so NaOH is the limiting reactant.

We calculate the produced moles of Na₂SO₃ using the moles of the limiting reactant:

• 0.82 mol NaOH * $\frac{1molNa_2SO_3}{2molNaOH}$ = 0.41 mol Na₂SO₃

Finally we convert Na₂SO₃ moles into grams, using its molar mass:

• 0.41 mol * 126.043 g/mol = 51.7 g Na₂SO₃