To balance the modified equation \(3H_2 + O_2 \rightarrow\) (products), let's first write down the unbalanced equation:
\[ 3H_2 + O_2 \rightarrow? \]
- The reactants consist of 3 hydrogen molecules (H2) and 1 oxygen molecule (O2).
- Each water molecule (H2O) consists of 2 hydrogen atoms and 1 oxygen atom.
Now, if we look to balance the equation, we can determine that:
- 3 molecules of \(H_2\) provide \(3 \times 2 = 6\) hydrogen atoms.
- To use those 6 hydrogen atoms, we would need \(3\) water molecules (since each water molecule contains 2 hydrogen atoms).
- Therefore, the balanced equation would be:
\[ 3H_2 + O_2 \rightarrow 3H_2O \]
Now looking at the statements provided:
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The equation will produce three molecules of water. - This is correct since the balanced equation shows that 3 molecules of \(H_2O\) are produced.
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It is not possible to produce water without leftover molecules. - This is incorrect because the reaction balances perfectly.
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Each oxygen molecule will combine with 2 hydrogen molecules with no leftover molecules. - This is partially correct in the context of a single \(O_2\) molecule, which does contribute to combining with 3 \(H_2\) molecules.
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The result of the equation will be the same. - This is incorrect because changing \(2H_2\) to \(3H_2\) changes the result.
The correct statement is: The equation will produce three molecules of water.