Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
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]
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]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.