Question

2h₂ + 1o₂ →→ 2h₂o
if you change the 2h₂ to 3h₂ molecules and balance the equation, identify which statement is correct?
(1 point)
○ the result of the equation will be the same.
○ each oxygen molecule will combine with 2 hydrogen molecules with no leftover molecules.
○ it is not possible to produce water without leftover molecules.
○ the equation will produce three molecules of water.

Explanation:

Step1: Analyze the original balanced equation

The original balanced equation \(2H_{2}+O_{2}
ightarrow2H_{2}O\) shows a \(2:1:2\) ratio of \(H_{2}:O_{2}:H_{2}O\).

Step2: Consider the new situation

If we change \(2H_{2}\) to \(3H_{2}\), let the balanced equation be \(3H_{2}+xO_{2}
ightarrow yH_{2}O\).
From hydrogen atom balance: \(3\times2 = 2y\), so \(y = 3\).
From oxygen atom balance: \(2x=3\), \(x = 1.5\). But we can't have a fraction of an \(O_{2}\) molecule in the context of whole - molecule reactions.
If we use whole - number coefficients for \(O_{2}\) (e.g., multiply the equation \(3H_{2}+1.5O_{2}
ightarrow3H_{2}O\) by 2 to get \(6H_{2}+3O_{2}
ightarrow6H_{2}O\)), but if we are limited to using non - fractional coefficients based on the initial change of \(2H_{2}\) to \(3H_{2}\) and not changing the nature of the reaction (i.e., still reacting \(H_{2}\) and \(O_{2}\) to form \(H_{2}O\)), there will be leftover molecules.
For example, if we have \(3H_{2}+O_{2}\), from the ratio in the original reaction (\(2H_{2}\) react with \(1O_{2}\)), \(3H_{2}\) will react with \(1.5O_{2}\). Since we can't have half - molecules in a simple reaction model (at the level of this problem), if we take \(1O_{2}\), \(2H_{2}\) will react (\(2H_{2}+O_{2}
ightarrow2H_{2}O\)) and \(1H_{2}\) will be left over.

Option A: The result (in terms of molecule amounts) is different.
Option B: With \(3H_{2}\), the ratio \(H_{2}:O_{2}\) is not \(2:1\) (required for no leftovers as in the original equation).
Option D: If we try to balance \(3H_{2}+O_{2}
ightarrow H_{2}O\) properly (using whole numbers for reactant molecules), we can't get exactly 3 water molecules without leftovers.

Answer:

C. It is not possible to produce water without leftover molecules.