To solve the problem of forming a final balanced chemical equation where [tex]\( HF \)[/tex] and [tex]\( O_2 \)[/tex] are the products formed through the reaction between [tex]\( H_2O \)[/tex] (water) and [tex]\( F_2 \)[/tex] (fluorine gas), we need to manipulate and combine the given intermediate chemical equations. Here are the given equations:
1. [tex]\( 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(l) \)[/tex]
2. [tex]\( H_2(g) + F_2(g) \rightarrow 2 HF(g) \)[/tex]
### Step-by-Step Solution:
1. Reverse the first equation:
This process indicates that the products will become reactants and vice versa.
[tex]\[ 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) \][/tex]
2. Multiply the second equation by 2:
This is done to ensure that the [tex]\( H_2 \)[/tex] produced in the first reverse equation can be canceled out with the [tex]\( H_2 \)[/tex] in the second equation when added.
[tex]\[ 2 \times \left( H_2(g) + F_2(g) \rightarrow 2 HF(g) \right) \][/tex]
Results in:
[tex]\[ 2 H_2(g) + 2 F_2(g) \rightarrow 4 HF(g) \][/tex]
3. Add the reversed first equation and the scaled second equation:
Now, we combine the two equations. On the left side of the overall equation, we have:
[tex]\[ 2 H_2O(l) + 2 F_2(g) \][/tex]
On the right side, we have:
[tex]\[ 2 H_2(g) + O_2(g) + 2 H_2(g) + 2 F_2(g) \rightarrow 2 H_2(g) + O_2(g) + 4 HF(g) \][/tex]
Notice that [tex]\( 2 H_2(g) \)[/tex] cancels out on both sides:
[tex]\[ 2 H_2O(l) + 2 F_2(g) \rightarrow O_2(g) + 4 HF(g) \][/tex]
Thus, the final balanced equation is:
[tex]\[ 2 H_2O(l) + 2 F_2(g) \rightarrow O_2(g) + 4 HF(g) \][/tex]
### Final Equation Details:
The coefficients and compounds in the final equation can be summarized as:
- Coefficients: [2, 2, 1, 4]
- Compounds: ["H2O(l)", "F2(g)", "O2(g)", "HF(g)"]
