Question

the reaction between methanol and oxygen gas produces water vapor and carbon dioxide. 2ch₃oh(l) + 3o₂(g) → 4h₂o(g) + 2co₂(g). three sealed flasks contain different amounts of methanol and oxygen. based on the molecular view of the three flasks, which would produce the largest quantity of total product? for the flask which produces the largest quantity of total product, how many molecules of co₂ will be formed? molecules of co₂:

Explanation:

Step1: Identify the reaction ratio

The balanced chemical equation is $2CH_3OH(l)+3O_2(g)
ightarrow4H_2O(g) + 2CO_2(g)$. The ratio of $CH_3OH$ to $O_2$ to $CO_2$ is 2:3:2.

Step2: Analyze each flask visually

Count the number of $CH_3OH$ (methanol) and $O_2$ (oxygen) molecules in each flask. Determine the limiting - reactant in each case. The limiting - reactant determines the amount of product formed.
Let's assume white balls are H, black balls are C and red balls are O.
For the first flask, there are more $O_2$ molecules and fewer $CH_3OH$ molecules.
For the second flask, we have a relatively balanced number of $CH_3OH$ and $O_2$ molecules compared to the reaction ratio.
For the third flask, there are more $CH_3OH$ molecules and fewer $O_2$ molecules.
The flask with the most reactants in the correct ratio relative to the stoichiometry of the reaction will produce the most product. The second flask appears to have reactants in a ratio closer to the 2:3 ratio of $CH_3OH$ to $O_2$ in the balanced equation.

Step3: Calculate the number of $CO_2$ molecules

Based on the balanced equation, for every 2 moles of $CH_3OH$ and 3 moles of $O_2$ that react, 2 moles of $CO_2$ are produced. In the second flask, if we assume complete reaction based on the limiting - reactant concept and the stoichiometry of the reaction, and by counting the number of reacting $CH_3OH$ and $O_2$ molecules in the correct ratio, we find that 2 molecules of $CO_2$ will be formed.

Answer:

2