Question

hydrogen ($ce{h_2}$) and oxygen ($ce{o_2}$) combine to make water ($ce{h_2o}$) in the following equation.
$ce{6h_2 + 3o_2 -> ?h_2o}$
how many water molecules would be produced?
(1 point)
○ five
○ six
○ three
○ four

Explanation:

Step1: Analyze Hydrogen Atoms

In the reactants, we have \(6\text{H}_2\), so the number of H atoms is \(6\times2 = 12\). Each \(\text{H}_2\text{O}\) has 2 H atoms. Let the number of \(\text{H}_2\text{O}\) be \(x\). Then \(2x=12\), so \(x = 6\) (we can also check oxygen: \(3\text{O}_2\) has \(3\times2 = 6\) O atoms, each \(\text{H}_2\text{O}\) has 1 O atom, so \(x = 6\) as well).

Step2: Confirm with Oxygen Atoms

Reactant oxygen: \(3\text{O}_2\) gives \(3\times2 = 6\) O atoms. Product water (\(\text{H}_2\text{O}\)) has 1 O per molecule. So number of \(\text{H}_2\text{O}\) molecules is \(6\) (since \(6\times1 = 6\) O atoms, matching reactant O).

Answer:

six (corresponding to the option with "six" as the text, e.g., if the option is labeled as, say, B. six, then B. six)