Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
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]
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.