lana balanced an equation so that the result was 2c₂h₃br + 5o₂ → 4co₂ + 2h₂o + 2hbr. which most likely…

lana balanced an equation so that the result was 2c₂h₃br + 5o₂ → 4co₂ + 2h₂o + 2hbr. which most likely represents the starting equation?\n2c₄h₃br + 5o₂ → 4co₂ + 2h₂o + 2hbr\nc₂h₃br + 5o₂ → 4co₂ + h₂o + 2hbr\nc₄h₃br + o₂ → co₂ + h₂o + hbr\nc₂h₃br + o₂ → co₂ + h₂o + hbr

lana balanced an equation so that the result was 2c₂h₃br + 5o₂ → 4co₂ + 2h₂o + 2hbr. which most likely represents the starting equation?\n2c₄h₃br + 5o₂ → 4co₂ + 2h₂o + 2hbr\nc₂h₃br + 5o₂ → 4co₂ + h₂o + 2hbr\nc₄h₃br + o₂ → co₂ + h₂o + hbr\nc₂h₃br + o₂ → co₂ + h₂o + hbr

Answer

Explanation:

Step1: Analyze balanced equation

The balanced equation is $2C_2H_3Br + 5O_2\rightarrow4CO_2 + 2H_2O+2HBr$. We need to find the un - balanced form.

Step2: Divide by coefficients

Dividing the balanced equation by 2 (since the coefficient of $C_2H_3Br$ is 2), we get $C_2H_3Br+\frac{5}{2}O_2\rightarrow2CO_2 + H_2O + HBr$. In an un - balanced equation we usually write whole - number coefficients for oxygen in the reactant side, but the key is the reactant and product species. The un - balanced equation before balancing must have been $C_2H_3Br + O_2\rightarrow CO_2 + H_2O + HBr$.

Answer:

D. $C_2H_3Br + O_2\rightarrow CO_2 + H_2O + HBr$