Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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 this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.