SO2 reacts with O2 to produce SO3. If 86.0 g of SO2 is placed in a reaction vessel along with excess oxygen gas, how many moles of SO2 remain when 50.0 g of SO3 have been formed?

0,72 moles of SO₂ remain

Explanation:

The reaction is:

2SO₂ + O₂ → 2SO₃

Where molecular mass of SO₂ is 64,066g/mol and of SO₃ is 80,066g/mol.

86,0g of SO₂ are:

86,0g × (1mol / 64,066g) = 1,34 moles of SO₂.

50,0g of SO₃ are:

50,0g × (1mol / 80,066g) = 0,62 moles of SO₃.

Now, as 2 moles of SO₂ produce 2 moles of SO₃, the moles of SO₂ that remain after the reaction are the initial moles of SO₂ - moles of SO₃:

1,34 moles - 0,62 moles =

0,72 moles of SO₂ remain

I hope it helps!