Explore IDNLearn.com's extensive Q&A database and find the answers you need. Discover detailed and accurate answers to your questions from our knowledgeable and dedicated community members.
Sagot :
To determine how many liters of [tex]\( C_2H_2 \)[/tex] react with 12.0 moles of [tex]\( O_2 \)[/tex] at standard temperature and pressure (STP), we will follow these steps:
1. Understanding the Stoichiometric Ratio:
According to the balanced chemical equation:
[tex]\[ 2 C_2H_2(g) + 5 O_2(g) \rightarrow 4 CO_2(g) + 2 H_2O(g) \][/tex]
The stoichiometric ratio between [tex]\( C_2H_2 \)[/tex] and [tex]\( O_2 \)[/tex] is 2:5. That means for every 2 moles of [tex]\( C_2H_2 \)[/tex], 5 moles of [tex]\( O_2 \)[/tex] are required.
2. Calculating Moles of [tex]\( C_2H_2 \)[/tex]:
Given that we have 12.0 moles of [tex]\( O_2 \)[/tex], we need to calculate how many moles of [tex]\( C_2H_2 \)[/tex] will react with it. Using the stoichiometric ratio [tex]\( \frac{2 \text{ moles } C_2H_2}{5 \text{ moles } O_2} \)[/tex]:
[tex]\[ \text{Moles of } C_2H_2 = 12.0 \text{ moles } O_2 \times \frac{2 \text{ moles } C_2H_2}{5 \text{ moles } O_2} = 4.8 \text{ moles } C_2H_2 \][/tex]
3. Converting Moles of [tex]\( C_2H_2 \)[/tex] to Liters:
At STP (Standard Temperature and Pressure: 0°C and 1 atm), one mole of any ideal gas occupies 22.4 liters. Thus, the volume of [tex]\( 4.8 \)[/tex] moles of [tex]\( C_2H_2 \)[/tex] can be calculated by:
[tex]\[ \text{Volume of } C_2H_2 = 4.8 \text{ moles } C_2H_2 \times 22.4 \text{ liters per mole} = 107.52 \text{ liters} \][/tex]
Therefore, the number of liters of [tex]\( C_2H_2 \)[/tex] that react with 12.0 moles of [tex]\( O_2 \)[/tex] at STP is 107.52 liters.
1. Understanding the Stoichiometric Ratio:
According to the balanced chemical equation:
[tex]\[ 2 C_2H_2(g) + 5 O_2(g) \rightarrow 4 CO_2(g) + 2 H_2O(g) \][/tex]
The stoichiometric ratio between [tex]\( C_2H_2 \)[/tex] and [tex]\( O_2 \)[/tex] is 2:5. That means for every 2 moles of [tex]\( C_2H_2 \)[/tex], 5 moles of [tex]\( O_2 \)[/tex] are required.
2. Calculating Moles of [tex]\( C_2H_2 \)[/tex]:
Given that we have 12.0 moles of [tex]\( O_2 \)[/tex], we need to calculate how many moles of [tex]\( C_2H_2 \)[/tex] will react with it. Using the stoichiometric ratio [tex]\( \frac{2 \text{ moles } C_2H_2}{5 \text{ moles } O_2} \)[/tex]:
[tex]\[ \text{Moles of } C_2H_2 = 12.0 \text{ moles } O_2 \times \frac{2 \text{ moles } C_2H_2}{5 \text{ moles } O_2} = 4.8 \text{ moles } C_2H_2 \][/tex]
3. Converting Moles of [tex]\( C_2H_2 \)[/tex] to Liters:
At STP (Standard Temperature and Pressure: 0°C and 1 atm), one mole of any ideal gas occupies 22.4 liters. Thus, the volume of [tex]\( 4.8 \)[/tex] moles of [tex]\( C_2H_2 \)[/tex] can be calculated by:
[tex]\[ \text{Volume of } C_2H_2 = 4.8 \text{ moles } C_2H_2 \times 22.4 \text{ liters per mole} = 107.52 \text{ liters} \][/tex]
Therefore, the number of liters of [tex]\( C_2H_2 \)[/tex] that react with 12.0 moles of [tex]\( O_2 \)[/tex] at STP is 107.52 liters.
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.