To balance the modified equation \(3H_2 + O_2 \rightarrow\), we need to adjust the number of water molecules produced.
Starting with the equation:
\[ 3H_2 + O_2 \rightarrow xH_2O \]
- Each water molecule (\(H_2O\)) requires 2 hydrogen atoms.
- With 3 molecules of \(H_2\), we have 6 hydrogen atoms available.
- Therefore, the number of water molecules produced (x) will be calculated as follows:
\[ 6 \text{ (hydrogens)} / 2 \text{ (hydrogens per water)} = 3 \text{ (water molecules)} \]
Now the balanced equation would be:
\[ 3H_2 + O_2 \rightarrow 3H_2O \]
This means three molecules of water are produced without any leftovers.
Therefore, the correct statement is:
Each oxygen molecule will combine with 2 hydrogen molecules with no leftover molecules.
This statement is true because in the balanced equation, one molecule of oxygen combines with three molecules of hydrogen to produce three molecules of water, and there are no leftover molecules.