Consider the reaction for the combustion of acetylene.
[tex]\[
2 \text{C}_2 \text{H}_2 (g) + 5 \text{O}_2 (g) \rightarrow 4 \text{CO}_2 (g) + 2 \text{H}_2 \text{O} (g)
\][/tex]

How many moles of [tex]\(\text{C}_2 \text{H}_2\)[/tex] react with 8.0 L of oxygen, assuming the reaction is at STP?

[tex]\[ \text{mol C}_2 \text{H}_2 \][/tex]



Answer :

To solve this question, we need to determine how many moles of acetylene [tex]\((C_2H_2)\)[/tex] react with 8.0 liters of oxygen [tex]\((O_2)\)[/tex], assuming the reaction occurs at standard temperature and pressure (STP).

1. Understanding the Reaction:
The balanced equation for the combustion of acetylene is:
[tex]\[ 2 C_2H_2(g) + 5 O_2(g) \rightarrow 4 CO_2(g) + 2 H_2O(g) \][/tex]
From this equation, we see that 2 moles of [tex]\(C_2H_2\)[/tex] react with 5 moles of [tex]\(O_2\)[/tex].

2. Molar Volume at STP:
The molar volume of any ideal gas at standard temperature and pressure (STP) is 22.4 liters per mole. This means 1 mole of gas occupies 22.4 liters.

3. Convert Volume of Oxygen to Moles:
First, we will convert the given volume of oxygen (8.0 liters) to moles using the molar volume at STP.
[tex]\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}} = \frac{8.0 \text{ L}}{22.4 \text{ L/mol}} \approx 0.357 \text{ moles} \][/tex]

4. Use Stoichiometry to Find Moles of Acetylene:
According to the balanced reaction, 5 moles of [tex]\(O_2\)[/tex] react with 2 moles of [tex]\(C_2H_2\)[/tex]. We can set up a proportion to find the moles of [tex]\(C_2H_2\)[/tex] that would react with the calculated moles of [tex]\(O_2\)[/tex].
[tex]\[ \frac{2 \text{ moles } C_2H_2}{5 \text{ moles } O_2} = \frac{x \text{ moles } C_2H_2}{0.357 \text{ moles } O_2} \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = \left( \frac{2}{5} \right) \times 0.357 \text{ moles O}_2 \][/tex]
[tex]\[ x \approx 0.143 \text{ moles } C_2H_2 \][/tex]

Thus, approximately 0.143 moles of [tex]\(C_2H_2\)[/tex] will react with 8.0 liters of [tex]\(O_2\)[/tex] at STP.