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Answer 1 · 2015-07-18T00:57:59

The decomposition will release 2.42 g of ${SO}_{2}$ $"SO"_2$ .

Your balanced equation is

$H_{2} {SO}_{3} \to H_{2} O + {SO}_{2}$ $"H"_2"SO"_3 → "H"_2"O" + "SO"_2$

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 $"SO"_2$ .

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