Let's solve the problem step-by-step.
The given chemical reaction is:
[tex]$2 H_2(g) + O_2(g) \rightarrow 2 H_2O(g)$[/tex]
This balanced equation tells us that:
- 2 moles of hydrogen gas ([tex]\(H_2\)[/tex]) react with 1 mole of oxygen gas ([tex]\(O_2\)[/tex])
- This reaction produces 2 moles of water ([tex]\(H_2O\)[/tex])
First, we need to determine the reaction stoichiometry, which, in this case, is:
- 1 mole of [tex]\(\text{O}_2\)[/tex] produces 2 moles of [tex]\(\text{H}_2O\)[/tex]
We are given:
- 3.0 moles of [tex]\(\text{O}_2(g)\)[/tex]
Using the stoichiometric ratio from the balanced equation, we calculate the moles of water ([tex]\(\text{H}_2O\)[/tex]) produced:
- For every 1 mole of [tex]\(\text{O}_2\)[/tex], 2 moles of [tex]\(\text{H}_2O\)[/tex] are formed.
Therefore, if 3.0 moles of [tex]\(\text{O}_2\)[/tex] react fully, the amount of produced water is:
[tex]$3.0 \text{ moles } \text{O}_2 \times \frac{2 \text{ moles } \text{H}_2O}{1 \text{ mole } \text{O}_2} = 6.0 \text{ moles } \text{H}_2O$[/tex]
Thus, 6.0 moles of [tex]\(\text{H}_2O(g)\)[/tex] will be made. Therefore, the correct choice is:
6.0 moles [tex]\(\text{H}_2O(g)\)[/tex]
