Question #6bcdf

1 Answer
Jul 18, 2015

Answer:

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#