Answer is: 116000 tons of sulfur dioxide was emitted.
Chemical reaction: S + O₂ → SO₂.
m(S) = m(coal) · ω(S).
m(S) = 2,9·10⁶ t · 0,02.
m(S) = 58000 t · 10⁶ g/t = 5,8·10¹⁰ g.
n(S) = m(S) ÷ M(S).
n(S) = 5,8·10¹⁰ g ÷ 32 g/mol.
n(S) = 1,8125·10⁹ mol.
From chemical reaction: n(S) : n(SO₂) = 1 : 1.
n(SO₂) = 1,8125·10⁹ mol.
m(SO₂) = 1,8125·10⁹ mol · 64 g/mol.
m(SO₂) = 1,16·10¹¹ g = 116000 t.
