To balance the chemical equation \( S_8 + O_2 \rightarrow SO_3 \), we need to ensure that the number of each type of atom is equal on both sides.
- Start with \( S_8 \) and \( O_2 \) on the left.
- The product is \( SO_3 \).
Let’s assign coefficients to each compound:
- Let \( a \) be the coefficient for \( S_8 \).
- Let \( b \) be the coefficient for \( O_2 \).
- Let \( c \) be the coefficient for \( SO_3 \).
The unbalanced equation can be rewritten as:
\[ a S_8 + b O_2 \rightarrow c SO_3 \]
From the left side, we have:
- \( 8a \) sulfur atoms (from \( S_8 \))
- \( 2b \) oxygen atoms (from \( O_2 \))
From the right side, we have:
- \( c \) sulfur atoms (from \( SO_3 \))
- \( 3c \) oxygen atoms (from \( SO_3 \))
Setting up the equations for balance:
-
For sulfur: \[ 8a = c \]
-
For oxygen: \[ 2b = 3c \]
Now, we start with \( a = 1 \) (which means we have 1 molecule of \( S_8 \)): \[ c = 8a = 8(1) = 8 \]
Now, substitute \( c \) into the oxygen equation: \[ 2b = 3(8) \implies 2b = 24 \implies b = 12 \]
So the coefficients are:
- \( a = 1 \) (for \( S_8 \))
- \( b = 12 \) (for \( O_2 \))
- \( c = 8 \) (for \( SO_3 \))
Putting this all together, the balanced equation is: \[ 1 S_8 + 12 O_2 \rightarrow 8 SO_3 \]
The coefficients to make both sides equal are: \[ 1, 12, 8 \]
Thus, the balanced equation is:
\[ S_8 + 12 O_2 \rightarrow 8 SO_3 \]