Answer :
To obtain the final chemical equation where \( HF \) and \( O_2 \) are the products from the reactions involving \( H_2O \) and \( F_2 \), we need to carefully manipulate the given intermediate chemical equations. Let's go through the necessary step-by-step modifications:
1. Given Intermediate Equations:
[tex]\[ \begin{array}{l} 1. \quad 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(l) \\ 2. \quad H_2(g) + F_2(g) \rightarrow 2 HF(g) \end{array} \][/tex]
2. Required Products: \( HF \) and \( O_2 \).
3. Reversing the First Equation: To obtain \( H_2O \) as a reactant, we need to reverse the first equation:
[tex]\[ 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) \][/tex]
4. Multiplying the Second Equation by 2: To balance the number of moles of \( H_2 \) and \( HF \), multiply the second equation by 2:
[tex]\[ 2 \left( H_2(g) + F_2(g) \rightarrow 2 HF(g) \right) \][/tex]
Simplifying, we get:
[tex]\[ 2 H_2(g) + 2 F_2(g) \rightarrow 4 HF(g) \][/tex]
5. Modified Equations: After reversing the first equation and multiplying the second equation by 2, the equations are:
[tex]\[ \begin{array}{l} 1. \quad 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) \\ 2. \quad 2 H_2(g) + 2 F_2(g) \rightarrow 4 HF(g) \end{array} \][/tex]
These steps give us the final modified intermediate equations, which are:
[tex]\[ \begin{array}{l} 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) \\ 2 H_2(g) + 2 F_2(g) \rightarrow 4 HF(g) \end{array} \][/tex]
Therefore, to derive the final chemical equation with [tex]\( HF \)[/tex] and [tex]\( O_2 \)[/tex] as products from reactions involving [tex]\( H_2O \)[/tex] and [tex]\( F_2 \)[/tex], you need to reverse the first equation and multiply the second equation by 2.
1. Given Intermediate Equations:
[tex]\[ \begin{array}{l} 1. \quad 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(l) \\ 2. \quad H_2(g) + F_2(g) \rightarrow 2 HF(g) \end{array} \][/tex]
2. Required Products: \( HF \) and \( O_2 \).
3. Reversing the First Equation: To obtain \( H_2O \) as a reactant, we need to reverse the first equation:
[tex]\[ 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) \][/tex]
4. Multiplying the Second Equation by 2: To balance the number of moles of \( H_2 \) and \( HF \), multiply the second equation by 2:
[tex]\[ 2 \left( H_2(g) + F_2(g) \rightarrow 2 HF(g) \right) \][/tex]
Simplifying, we get:
[tex]\[ 2 H_2(g) + 2 F_2(g) \rightarrow 4 HF(g) \][/tex]
5. Modified Equations: After reversing the first equation and multiplying the second equation by 2, the equations are:
[tex]\[ \begin{array}{l} 1. \quad 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) \\ 2. \quad 2 H_2(g) + 2 F_2(g) \rightarrow 4 HF(g) \end{array} \][/tex]
These steps give us the final modified intermediate equations, which are:
[tex]\[ \begin{array}{l} 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) \\ 2 H_2(g) + 2 F_2(g) \rightarrow 4 HF(g) \end{array} \][/tex]
Therefore, to derive the final chemical equation with [tex]\( HF \)[/tex] and [tex]\( O_2 \)[/tex] as products from reactions involving [tex]\( H_2O \)[/tex] and [tex]\( F_2 \)[/tex], you need to reverse the first equation and multiply the second equation by 2.
