Question

use the equation to answer the question. ch₄ + 2o₂ → co₂ + 2h₂o if you change the 2 in front of 2o₂ to a 3, what will be the change in the results on the right side of the equation? (1 point) nothing changes in the equation there is an extra o₂ molecule left over there are now 3 h₂o molecules there are now 3 co₂ molecules

Explanation:

Step1: Analyze original equation

The original balanced equation is $CH_4 + 2O_2
ightarrow CO_2+2H_2O$. Here, 1 mole of $CH_4$ reacts with 2 moles of $O_2$ to produce 1 mole of $CO_2$ and 2 moles of $H_2O$.

Step2: Consider new reactant amount

When we change the coefficient of $O_2$ from 2 to 3 ($CH_4 + 3O_2$), the $CH_4$ is still 1 mole. For complete combustion of 1 mole of $CH_4$, only 2 moles of $O_2$ are required. So, $3 - 2=1$ mole of $O_2$ will be left over.

Answer:

There is an extra $O_2$ molecule left over