Answer :
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.
