Question

Question #6bcdf

Answer 1

The decomposition will release 2.42 g of `"SO"_2`.

Explanation:

Your balanced equation is

`"H"_2"SO"_3 → "H"_2"O" + "SO"_2`

Step 1. Calculate the moles of sulfurous acid.

`3.10 cancel("g H₂SO₃") × ("1 mol H"_2"SO"_3)/(82.08 cancel("g H₂SO₃")) = "0.037 77 mol H"_2"SO"_3`

Step 2. Calculate the moles of `"SO"_2`.

`0.037 77 cancel("mol H₂SO₃") × ("1 mol SO"_2)/(1 cancel("mol H₂SO₃")) = "0.037 77 mol SO"_2`

Step 3. Calculate the mass of `"SO"_2`.

`0.037 77 cancel("mol SO₂") × ("64.06 g SO"_2)/(1 cancel("mol SO₂")) = "2.42 g SO"_2`