Let's start by balancing the modified equation:
If we change \(2H_2\) to \(3H_2\), the new unbalanced equation is:
\[ 3H_2 + O_2 \rightarrow \text{?} H_2O \]
To balance this equation, we need to determine how many \(H_2O\) molecules will be produced.
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Balance Hydrogen: Since we have 3 \(H_2\), we have a total of \(3 \times 2 = 6\) hydrogen atoms on the reactant side. Therefore, to balance hydrogen, we need 6 hydrogen atoms on the product side, which means we need: \[ \frac{6}{2} = 3 \text{ molecules of } H_2O \]
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Balance Oxygen: We produce 3 \(H_2O\), which contains \(3 \times 1 = 3\) oxygen atoms. To balance the oxygen in the reaction, we need to adjust the \(O_2\): \[ O_2 \text{ required for 3 } H_2O = \frac{3}{2}O_2 \]
Putting it all together, the balanced equation is:
\[ 3H_2 + \frac{3}{2}O_2 \rightarrow 3H_2O \]
This shows that when you start with 3 \(H_2\), it will produce 3 \(H_2O\) without any leftover molecules of hydrogen.
Now let's evaluate the given statements:
- The equation will produce three molecules of water. - This is correct.
- The result of the equation will be the same. - This is incorrect because the amount of water produced has changed.
- It is not possible to produce water without leftover molecules. - This is incorrect since the balanced equation produces water without leftover molecules.
- Each oxygen molecule will combine with 2 hydrogen molecules with no leftover molecules. - This is also incorrect. In this case, we need 1.5 \(O_2\) (which cannot be taken as a whole number for molecules) to combine with 3 \(H_2\).
Thus, the correct response is:
The equation will produce three molecules of water.