After the following equation is properly balanced, what is the coefficient in front of O₂?
S₈(s) + O₂(g) $\rightarrow$ SO₃(g)

Question

After the following equation is properly balanced, what is the coefficient in front of O₂? S₈(s) + O₂(g) $\rightarrow$SO₃(g)

Quizplus · Accepted Answer

To balance this equation, we need to ensure that the number of each type of atom is the same on both sides of the equation. Starting with sulfur (S), we have: - 18 S atoms on the left (in S₈) - 13 S atoms on the right (in SO₃) To balance the sulfur atoms, we need to multiply the sulfur dioxide (SO2) on the left by 2: - 2 SO2 + O₂(g) F1F1F1S1F1F1F10 SO₃(g) + 2 S₈(s) Now, we have: - 36 S atoms on the left (in 2 S₈) - 13 S atoms on the right (in SO₃) To balance the oxygen (O) atoms, we can multiply the fluorine (F2) on the right by 6: - 2 SO2 + O₂(g) + 6 F2S1F1F1F10 SO₃(g) + 2 S₈(s) Now, we have: - 36 S atoms on the left (in 2 S₈) - 13 S atoms on the right (in SO₃) - 36 O atoms on the left (in 2 SO2 and S₈) - 78 O atoms on the right (in SO₃ and F1F1F10) Finally, to balance the osmium (Os) and boron (B) atoms, we need to add coefficients of 12 and 1, respectively, in front of the osmium boride on the left: - 2 SO2 + 12 O₂(g) + 6 F2S1F1F1F10 SO₃(g) + 2 S₈(s) Now, we have: - 36 S atoms on the left (in 2 S₈) - 13 S atoms on the right (in SO₃) - 36 O atoms on the left (in 2 SO2 and S₈) - 78 O atoms on the right (in SO₃ and F1F1F10) - 12 Os atoms on the left (in 12 O₂) - 12 Os atoms on the right (in SO₃) - 12 B atoms on the left (in 12 O₂) - 12 B atoms on the right (in SO₃) Therefore, the coefficient in front of O₂ is 12.

After the Following Equation Is Properly Balanced, What Is the Coefficient