To determine how many moles of water ([tex]\(H_2O\)[/tex]) can be formed from 2.4 moles of methane ([tex]\(CH_4\)[/tex]) and 6.5 moles of oxygen ([tex]\(O_2\)[/tex]), we need to follow these steps:
1. Write the balanced chemical equation:
[tex]\[
CH_4 + 2O_2 \rightarrow 2H_2O + CO_2
\][/tex]
This indicates that 1 mole of [tex]\(CH_4\)[/tex] reacts with 2 moles of [tex]\(O_2\)[/tex] to produce 2 moles of [tex]\(H_2O\)[/tex].
2. Determine the moles of [tex]\(H_2O\)[/tex] each reactant can produce:
- For [tex]\(CH_4\)[/tex]:
[tex]\[
1 \text{ mole of } CH_4 \text{ produces } 2 \text{ moles of } H_2O
\][/tex]
So, for 2.4 moles of [tex]\(CH_4\)[/tex]:
[tex]\[
2.4 \text{ moles } CH_4 \times \frac{2 \text{ moles } H_2O}{1 \text{ mole } CH_4} = 4.8 \text{ moles } H_2O
\][/tex]
- For [tex]\(O_2\)[/tex]:
[tex]\[
2 \text{ moles of } O_2 \text{ produce } 2 \text{ moles of } H_2O
\][/tex]
Thus, for 6.5 moles of [tex]\(O_2\)[/tex]:
[tex]\[
6.5 \text{ moles } O_2 \times \frac{2 \text{ moles } H_2O}{2 \text{ moles } O_2} = 6.5 \text{ moles } H_2O
\][/tex]
3. Identify the limiting reactant:
The limiting reactant is the one that produces the lesser amount of product (in this case, [tex]\(H_2O\)[/tex]). Therefore, we compare the moles of [tex]\(H_2O\)[/tex] that can be produced by each reactant:
[tex]\[
\text{Moles of } H_2O \text{ from } CH_4: 4.8 \text{ moles }
\][/tex]
[tex]\[
\text{Moles of } H_2O \text{ from } O_2: 6.5 \text{ moles }
\][/tex]
Since 4.8 moles is the lesser amount, [tex]\(CH_4\)[/tex] is the limiting reactant.
4. Calculate the moles of [tex]\(H_2O\)[/tex] formed:
The number of moles of [tex]\(H_2O\)[/tex] formed is dictated by the limiting reactant, which is [tex]\(CH_4\)[/tex].
Therefore, the number of moles of [tex]\(H_2O\)[/tex] that can be formed is:
[tex]\[
\boxed{4.8 \text{ moles}}
\][/tex]
