Consider the following equation:
[tex]\[ 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(g) \][/tex]

If 3.0 moles of [tex]\( O_2(g) \)[/tex] react fully with hydrogen, how many moles of water will be made?

A. 6.5 moles [tex]\( H_2O(g) \)[/tex]
B. 3.0 moles [tex]\( H_2O(g) \)[/tex]
C. 1.5 moles [tex]\( H_2O(g) \)[/tex]
D. 6.0 moles [tex]\( H_2O(g) \)[/tex]



Answer :

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]

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