To balance the modified equation with 3H₂ (instead of 2H₂), we start with:
3H₂ + O₂ → ?
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For 3 molecules of H₂, which is \(3 \times 2 = 6\) hydrogen atoms, we need 3 molecules of water (H₂O) to balance the hydrogen.
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The balanced equation would then be:
\[ 3H₂ + \frac{3}{2}O₂ → 3H₂O \]
However, since we typically express equations using whole numbers, we can multiply the entire equation by 2 to eliminate the fraction:
\[ 6H₂ + 3O₂ → 6H₂O \]
Now examining the statements:
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It is not possible to produce water without leftover molecules. - This is not true since the equation is balanced, and all the reactants are consumed.
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The equation will produce three molecules of water. - This is true if referring to the original unbalanced equation with 3H₂.
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The result of the equation will be the same. - This is a bit ambiguous, but in context, this can imply that water is still produced, which is true but under different stoichiometric amounts.
The most straightforward correct statement is:
- The equation will produce three molecules of water.