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Sagot :
To determine the final chemical equation, we need to manipulate the given intermediate equations appropriately. Here is a step-by-step solution:
1. Identify the Given Equations:
[tex]\[ \text{(Equation 1)} \ 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(l) \][/tex]
[tex]\[ \text{(Equation 2)} \ H_2(g) + F_2(g) \rightarrow 2 HF(g) \][/tex]
2. Alter the Second Equation:
Multiply the second equation by 2 to ensure the stoichiometric balance for [tex]\( H_2(g) \)[/tex]:
[tex]\[ 2 \times \left( H_2(g) + F_2(g) \rightarrow 2 HF(g) \right) \][/tex]
This gives us:
[tex]\[ 2 H_2(g) + 2 F_2(g) \rightarrow 4 HF(g) \][/tex]
3. Reverse the First Equation:
Reversing the first equation will change the products to reactants and vice versa:
[tex]\[ 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) \][/tex]
4. Combine the Altered Equations:
Now we have:
[tex]\[ 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) \][/tex]
and
[tex]\[ 2 H_2(g) + 2 F_2(g) \rightarrow 4 HF(g) \][/tex]
Combining these two altered equations, we align the [tex]\( H_2(g) \)[/tex] on both sides, leading to:
[tex]\[ 2 H_2O(l) + 2 F_2(g) \rightarrow 2 H_2(g) + O_2(g) + 4 HF(g) - 2 H_2(g) \][/tex]
The [tex]\( 2 H_2(g) \)[/tex] terms cancel each other out:
[tex]\[ 2 H_2O(l) + 2 F_2(g) \rightarrow 4 HF(g) + O_2(g) \][/tex]
5. Write the Final Equation:
The final chemical equation is:
[tex]\[ 2 H_2O(l) + 2 F_2(g) \rightarrow 4 HF(g) + O_2(g) \][/tex]
In summary, by carefully multiplying, reversing, and combining the intermediate equations, we arrive at the final balanced equation where [tex]\( HF \)[/tex] and [tex]\( O_2 \)[/tex] are the products formed from the reaction between [tex]\( H_2O \)[/tex] and [tex]\( F_2 \)[/tex]. The final equation is:
[tex]\[ 2 H_2O(l) + 2 F_2(g) \rightarrow 4 HF(g) + O_2(g) \][/tex]
1. Identify the Given Equations:
[tex]\[ \text{(Equation 1)} \ 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(l) \][/tex]
[tex]\[ \text{(Equation 2)} \ H_2(g) + F_2(g) \rightarrow 2 HF(g) \][/tex]
2. Alter the Second Equation:
Multiply the second equation by 2 to ensure the stoichiometric balance for [tex]\( H_2(g) \)[/tex]:
[tex]\[ 2 \times \left( H_2(g) + F_2(g) \rightarrow 2 HF(g) \right) \][/tex]
This gives us:
[tex]\[ 2 H_2(g) + 2 F_2(g) \rightarrow 4 HF(g) \][/tex]
3. Reverse the First Equation:
Reversing the first equation will change the products to reactants and vice versa:
[tex]\[ 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) \][/tex]
4. Combine the Altered Equations:
Now we have:
[tex]\[ 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) \][/tex]
and
[tex]\[ 2 H_2(g) + 2 F_2(g) \rightarrow 4 HF(g) \][/tex]
Combining these two altered equations, we align the [tex]\( H_2(g) \)[/tex] on both sides, leading to:
[tex]\[ 2 H_2O(l) + 2 F_2(g) \rightarrow 2 H_2(g) + O_2(g) + 4 HF(g) - 2 H_2(g) \][/tex]
The [tex]\( 2 H_2(g) \)[/tex] terms cancel each other out:
[tex]\[ 2 H_2O(l) + 2 F_2(g) \rightarrow 4 HF(g) + O_2(g) \][/tex]
5. Write the Final Equation:
The final chemical equation is:
[tex]\[ 2 H_2O(l) + 2 F_2(g) \rightarrow 4 HF(g) + O_2(g) \][/tex]
In summary, by carefully multiplying, reversing, and combining the intermediate equations, we arrive at the final balanced equation where [tex]\( HF \)[/tex] and [tex]\( O_2 \)[/tex] are the products formed from the reaction between [tex]\( H_2O \)[/tex] and [tex]\( F_2 \)[/tex]. The final equation is:
[tex]\[ 2 H_2O(l) + 2 F_2(g) \rightarrow 4 HF(g) + O_2(g) \][/tex]
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