Sure, let's balance the given chemical reaction step-by-step:
The unbalanced chemical equation is:
[tex]\[ C_4H_{10}(g) + O_2(g) \rightarrow CO_2(g) + H_2O(l) \][/tex]
Step 1: Balance the carbon atoms.
On the left side, there are 4 carbon atoms in [tex]\( C_4H_{10} \)[/tex]. Therefore, we need 4 molecules of [tex]\( CO_2 \)[/tex] to have 4 carbon atoms on the right side:
[tex]\[ C_4H_{10}(g) + O_2(g) \rightarrow 4CO_2(g) + H_2O(l) \][/tex]
Step 2: Balance the hydrogen atoms.
On the left side, there are 10 hydrogen atoms in [tex]\( C_4H_{10} \)[/tex]. Therefore, we need 5 molecules of [tex]\( H_2O \)[/tex] to have 10 hydrogen atoms on the right side:
[tex]\[ C_4H_{10}(g) + O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l) \][/tex]
Step 3: Balance the oxygen atoms.
Now we count the oxygen atoms on the right side. There are:
- 4 molecules of [tex]\( CO_2 \)[/tex], each containing 2 oxygen atoms, so [tex]\( 4 \times 2 = 8 \)[/tex] oxygen atoms from [tex]\( CO_2 \)[/tex].
- 5 molecules of [tex]\( H_2O \)[/tex], each containing 1 oxygen atom, so [tex]\( 5 \times 1 = 5 \)[/tex] oxygen atoms from [tex]\( H_2O \)[/tex].
In total, there are [tex]\( 8 + 5 = 13 \)[/tex] oxygen atoms on the right side.
To balance the oxygen atoms on the left side, each [tex]\( O_2 \)[/tex] molecule contains 2 oxygen atoms. Therefore, we need [tex]\( \frac{13}{2} = 6.5 \)[/tex] molecules of [tex]\( O_2 \)[/tex].
Since we cannot have a fraction of a molecule in a balanced equation, we multiply all coefficients to make them whole numbers. The lowest common factor here is 2. So, we multiply every coefficient by 2:
[tex]\[ 2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(l) \][/tex]
Thus, the balanced equation is:
[tex]\[ 2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(l) \][/tex]
