IDNLearn.com is your go-to resource for finding precise and accurate answers. Our experts provide accurate and detailed responses to help you navigate any topic or issue with confidence.
Sagot :
To solve the problem of determining the amount of moles of products and remaining reactants after a chemical reaction, we will follow these steps:
1. Identify the Initial Quantities:
- Initially, we have 0.280 mol of [tex]\( O_2 \)[/tex].
- Initially, we have 0.871 mol of [tex]\( H_2 \)[/tex].
2. Determine the Limiting Reactant:
- The balanced chemical equation is:
[tex]\[ O_2(g) + 2 H_2(g) \longrightarrow 2 H_2O(l) \][/tex]
- According to the equation, 1 mole of [tex]\( O_2 \)[/tex] reacts with 2 moles of [tex]\( H_2 \)[/tex]. Therefore, to fully react 0.280 mol of [tex]\( O_2 \)[/tex], we need:
[tex]\[ 0.280 \text{ mol } O_2 \times 2 = 0.560 \text{ mol } H_2 \][/tex]
- We compare this with the available [tex]\( H_2 \)[/tex]:
[tex]\[ 0.871 \text{ mol } H_2 \geq 0.560 \text{ mol } H_2 \][/tex]
- Since we have enough [tex]\( H_2 \)[/tex], [tex]\( O_2 \)[/tex] is the limiting reactant.
3. Calculate the Amount of Each Substance After the Reaction:
- All 0.280 mol of [tex]\( O_2 \)[/tex] will be consumed. Therefore:
[tex]\[ \text{Moles of } O_2 \text{ used} = 0.280 \text{ mol} \][/tex]
- Accordingly, the amount of [tex]\( H_2 \)[/tex] used is:
[tex]\[ \text{Moles of } H_2 \text{ used} = 0.280 \text{ mol} \times 2 = 0.560 \text{ mol} \][/tex]
- The remaining [tex]\( H_2 \)[/tex] is:
[tex]\[ 0.871 \text{ mol} - 0.560 \text{ mol} = 0.311 \text{ mol} \][/tex]
- The moles of [tex]\( H_2O \)[/tex] produced are:
[tex]\[ 2 \times 0.280 \text{ mol} = 0.560 \text{ mol} \][/tex]
4. Summarize the Results:
- Moles of water ([tex]\( H_2O \)[/tex]) produced: 0.560 mol
- Moles of remaining hydrogen ([tex]\( H_2 \)[/tex]): 0.311 mol
- Moles of remaining oxygen ([tex]\( O_2 \)[/tex]): 0.000 mol
Therefore:
[tex]\[ H_2 O : 0.560 \text{ mol} \][/tex]
[tex]\[ H_2 : 0.311 \text{ mol} \][/tex]
[tex]\[ O_2 : 0.000 \text{ mol} \][/tex]
1. Identify the Initial Quantities:
- Initially, we have 0.280 mol of [tex]\( O_2 \)[/tex].
- Initially, we have 0.871 mol of [tex]\( H_2 \)[/tex].
2. Determine the Limiting Reactant:
- The balanced chemical equation is:
[tex]\[ O_2(g) + 2 H_2(g) \longrightarrow 2 H_2O(l) \][/tex]
- According to the equation, 1 mole of [tex]\( O_2 \)[/tex] reacts with 2 moles of [tex]\( H_2 \)[/tex]. Therefore, to fully react 0.280 mol of [tex]\( O_2 \)[/tex], we need:
[tex]\[ 0.280 \text{ mol } O_2 \times 2 = 0.560 \text{ mol } H_2 \][/tex]
- We compare this with the available [tex]\( H_2 \)[/tex]:
[tex]\[ 0.871 \text{ mol } H_2 \geq 0.560 \text{ mol } H_2 \][/tex]
- Since we have enough [tex]\( H_2 \)[/tex], [tex]\( O_2 \)[/tex] is the limiting reactant.
3. Calculate the Amount of Each Substance After the Reaction:
- All 0.280 mol of [tex]\( O_2 \)[/tex] will be consumed. Therefore:
[tex]\[ \text{Moles of } O_2 \text{ used} = 0.280 \text{ mol} \][/tex]
- Accordingly, the amount of [tex]\( H_2 \)[/tex] used is:
[tex]\[ \text{Moles of } H_2 \text{ used} = 0.280 \text{ mol} \times 2 = 0.560 \text{ mol} \][/tex]
- The remaining [tex]\( H_2 \)[/tex] is:
[tex]\[ 0.871 \text{ mol} - 0.560 \text{ mol} = 0.311 \text{ mol} \][/tex]
- The moles of [tex]\( H_2O \)[/tex] produced are:
[tex]\[ 2 \times 0.280 \text{ mol} = 0.560 \text{ mol} \][/tex]
4. Summarize the Results:
- Moles of water ([tex]\( H_2O \)[/tex]) produced: 0.560 mol
- Moles of remaining hydrogen ([tex]\( H_2 \)[/tex]): 0.311 mol
- Moles of remaining oxygen ([tex]\( O_2 \)[/tex]): 0.000 mol
Therefore:
[tex]\[ H_2 O : 0.560 \text{ mol} \][/tex]
[tex]\[ H_2 : 0.311 \text{ mol} \][/tex]
[tex]\[ O_2 : 0.000 \text{ mol} \][/tex]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.
