To balance the modified equation with 3H₂, the equation would look like this:
\[ 3H_2 + O_2 \rightarrow 2H_2O \]
Balancing the equation:
- 3 molecules of H₂ provide 6 hydrogen atoms.
- Each molecule of O₂ provides 2 oxygen atoms, which can combine with the 6 hydrogen atoms to form 3 molecules of water (H₂O).
Thus, the balanced equation is:
\[ 3H_2 + 1.5O_2 \rightarrow 3H_2O \]
However, if you express the oxygen in whole numbers, it can be represented as:
\[ 6H_2 + 3O_2 \rightarrow 6H_2O \]
Now let's evaluate the statements:
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The equation will produce three molecules of water. - This statement is incorrect; it produces three molecules of water if 3H₂ is used, or six molecules if balanced correctly with whole coefficients.
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The result of the equation will be the same. - This is vague. If you're changing the reactants to 3H₂, the products change accordingly.
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It is not possible to produce water without leftover molecules. - This statement is incorrect; in the current balancing, it is possible to balance without leftovers.
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Each oxygen molecule will combine with 2 hydrogen molecules with no leftover molecules. - This statement is also incorrect; when using 3H₂, each O₂ molecule combines with 3 H₂ molecules in the correctly balanced equation.
Given this analysis, none of the provided statements perfectly capture the outcome of changing 2H₂ to 3H₂ and balancing the equation. That said, the first statement's confusion about how many water molecules are produced is important, and the wording could lead to misunderstanding the actual results of stoichiometry.
However, if asked to choose one, "Each oxygen molecule will combine with 2 hydrogen molecules with no leftover molecules." is often used in standard approaches for interpreting water formation, but it does not reflect the balance when 3H2 is used. The better balanced form indicates that with a proper understanding, it's possible to have exact ratios without leftovers.