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To determine how many moles of [tex]\( C_2H_2 \)[/tex] react with 8.0 liters of oxygen ([tex]\( O_2 \)[/tex]), we start with the provided balanced chemical equation:
[tex]\[ 2 C_2H_2(g) + 5 O_2(g) \rightarrow 4 CO_2(g) + 2 H_2O(g) \][/tex]
Here's the step-by-step solution:
1. Determine moles of [tex]\( O_2 \)[/tex]:
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. We can use this fact to convert the volume of [tex]\( O_2 \)[/tex] to moles.
Given:
[tex]\[ \text{Volume of } O_2 = 8.0 \text{ liters} \][/tex]
At STP:
[tex]\[ \text{Volume of 1 mole of gas} = 22.4 \text{ liters} \][/tex]
Calculate the moles of [tex]\( O_2 \)[/tex]:
[tex]\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Volume of 1 mole of gas}} \][/tex]
[tex]\[ \text{Moles of } O_2 = \frac{8.0 \text{ liters}}{22.4 \text{ liters/mole}} = 0.35714285714285715 \text{ moles} \][/tex]
2. Use the balanced equation to find the stoichiometric relationship:
According to the balanced equation:
[tex]\[ 2 \text{ moles of } C_2H_2 \text{ react with } 5 \text{ moles of } O_2 \][/tex]
We need to find the moles of [tex]\( C_2H_2 \)[/tex] that would react with the calculated moles of [tex]\( O_2 \)[/tex]:
Using the stoichiometric ratio:
[tex]\[ \text{Moles of } C_2H_2 = \left( \frac{2 \text{ moles } C_2H_2}{5 \text{ moles } O_2} \right) \times \text{Moles of } O_2 \][/tex]
[tex]\[ \text{Moles of } C_2H_2 = \left( \frac{2}{5} \right) \times 0.35714285714285715 \text{ moles} \][/tex]
3. Calculate the moles of [tex]\( C_2H_2 \)[/tex]:
[tex]\[ \text{Moles of } C_2H_2 = 0.14285714285714288 \text{ moles} \][/tex]
Therefore, the number of moles of [tex]\( C_2H_2 \)[/tex] that react with 8.0 liters of oxygen at STP is [tex]\( \boxed{0.14285714285714288 \text{ moles}} \)[/tex].
[tex]\[ 2 C_2H_2(g) + 5 O_2(g) \rightarrow 4 CO_2(g) + 2 H_2O(g) \][/tex]
Here's the step-by-step solution:
1. Determine moles of [tex]\( O_2 \)[/tex]:
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. We can use this fact to convert the volume of [tex]\( O_2 \)[/tex] to moles.
Given:
[tex]\[ \text{Volume of } O_2 = 8.0 \text{ liters} \][/tex]
At STP:
[tex]\[ \text{Volume of 1 mole of gas} = 22.4 \text{ liters} \][/tex]
Calculate the moles of [tex]\( O_2 \)[/tex]:
[tex]\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Volume of 1 mole of gas}} \][/tex]
[tex]\[ \text{Moles of } O_2 = \frac{8.0 \text{ liters}}{22.4 \text{ liters/mole}} = 0.35714285714285715 \text{ moles} \][/tex]
2. Use the balanced equation to find the stoichiometric relationship:
According to the balanced equation:
[tex]\[ 2 \text{ moles of } C_2H_2 \text{ react with } 5 \text{ moles of } O_2 \][/tex]
We need to find the moles of [tex]\( C_2H_2 \)[/tex] that would react with the calculated moles of [tex]\( O_2 \)[/tex]:
Using the stoichiometric ratio:
[tex]\[ \text{Moles of } C_2H_2 = \left( \frac{2 \text{ moles } C_2H_2}{5 \text{ moles } O_2} \right) \times \text{Moles of } O_2 \][/tex]
[tex]\[ \text{Moles of } C_2H_2 = \left( \frac{2}{5} \right) \times 0.35714285714285715 \text{ moles} \][/tex]
3. Calculate the moles of [tex]\( C_2H_2 \)[/tex]:
[tex]\[ \text{Moles of } C_2H_2 = 0.14285714285714288 \text{ moles} \][/tex]
Therefore, the number of moles of [tex]\( C_2H_2 \)[/tex] that react with 8.0 liters of oxygen at STP is [tex]\( \boxed{0.14285714285714288 \text{ moles}} \)[/tex].
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