consider the following intermediate chemical equations. 2h₂(g)+o₂(g)→2h₂o(l) h₂(g)+f₂(g)→2hf(g) in the final…
consider the following intermediate chemical equations. 2h₂(g)+o₂(g)→2h₂o(l) h₂(g)+f₂(g)→2hf(g) in the final chemical equation, hf and o₂ are the products that are formed through the reaction between h₂o and f₂. before you can add these intermediate chemical equations, you need to alter them by multiplying the second equation by 2 and reversing the first equation. first equation by 2 and reversing it. first equation by (1/2) and reversing the second equation. second equation by 2 and reversing it.
Answer
Explanation:
Step1: Analizar la reacción final
La reacción final es $2H_2O + 2F_2\rightarrow 4HF+O_2$.
Step2: Analizar la primera ecuación
La primera ecuación es $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$. Para que $H_2O$ sea reactivo, debemos revertirla: $2H_2O(l)\rightarrow 2H_2(g)+O_2(g)$.
Step3: Analizar la segunda ecuación
La segunda ecuación es $H_2(g)+F_2(g)\rightarrow 2HF(g)$. Para tener $4HF$ en el producto y equilibrar con la reacción final, debemos multiplicarla por 2: $2H_2(g)+2F_2(g)\rightarrow 4HF(g)$.
Answer:
A. second equation by 2 and reversing the first equation.