To balance the equation after changing \(2H_2\) to \(3H_2\), the new reaction would be:
\[ 3H_2 + \text{O}_2 \rightarrow ? \]
To balance:
- The left side now has 6 hydrogen atoms (from \(3H_2\)).
- To balance the hydrogen on the right side, we need 3 water molecules (\(3H_2O\)), which will give us 6 hydrogen atoms.
Thus, the balanced equation is:
\[ 3H_2 + \text{O}_2 \rightarrow 3H_2O \]
Next, we need to balance the oxygen:
- The right side has \(3O\) from \(3H_2O\).
- To balance this, we need 1.5 \(O_2\) on the left (1.5 molecules of \(\text{O}_2\) gives us 3 oxygen atoms). However, in terms of whole molecules, we can multiply the entire equation by 2 to avoid fractions:
\[ 6H_2 + 3O_2 \rightarrow 6H_2O \]
Now that we have the balanced equation, we can evaluate the statements.
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The equation will produce three molecules of water.
- This is incorrect; it actually produces six molecules of water.
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The result of the equation will be the same.
- This is ambiguous. If by "result" we mean the chemical product, it's the same (water), but the amount produced differs when comparing \(2H_2\) to \(3H_2\).
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Each oxygen molecule will combine with 2 hydrogen molecules with no leftover molecules.
- This is incorrect as \(3H_2\) would require 1.5 \(O_2\) to fully react without leftovers.
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It is not possible to produce water without leftover molecules.
- This is the most accurate statement, especially since with \(3H_2\) you can't have whole O2 molecules reacting without leftovers (unless adjusting for fractions, but within the context of the question, we are looking at whole molecules).
Given these evaluations, the correct response is:
It is not possible to produce water without leftover molecules.