xCH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)O
Balancing the equation, we have
C = x (on the left) and 2 (on the right)
⇒ x = 2
2CH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)O
On the left, H = 8 and on the right = 2z
⇒ 2z = 8 ; z = 4
2CH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + 4H\(_2\)O
On the right, O = 4 + 4 = 8.
On the left, O = 2 + 2y
⇒ 2 + 2y = 8
2y = 6 ; y = 3
x, y, z = 2, 3, 4.