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.

51.7 g

Explanation:

The reaction that takes place is:

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

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:

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

no podemos ver la

explanation:

no te podemos ayudar: /