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Given the reaction:

[tex]\[ 2 C_2H_2(g) + 5 O_2(g) \rightarrow 4 CO_2(g) + 2 H_2O(g) \][/tex]

How many liters of [tex]\( C_2H_2 \)[/tex] react with 12.0 mol [tex]\( O_2 \)[/tex], assuming the reaction is at STP?

[tex]\[
\begin{array}{c|c|c}
12.0 \text{ mol } O_2 & \frac{2 \text{ mol } C_2H_2}{5 \text{ mol } O_2} & \frac{22.4 \text{ L } C_2H_2}{1 \text{ mol } C_2H_2} \\
\hline
& & \\
\end{array}
= [?] \text{ L } C_2H_2
\][/tex]

---

Response:

Sagot :

GinnyAnsweridn
To solve this problem, we need to determine how many liters of acetylene ([tex]\( \text{C}_2\text{H}_2 \)[/tex]) react with 12.0 moles of oxygen ([tex]\( \text{O}_2 \)[/tex]) given the balanced chemical equation:

[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]

Here's a detailed, step-by-step solution:

1. Understanding the Mole Ratio:
The balanced chemical equation shows that 2 moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] react with 5 moles of [tex]\( \text{O}_2 \)[/tex]. Therefore, we can write the mole ratio:
[tex]\[ \frac{2 \text{ moles } \text{C}_2\text{H}_2}{5 \text{ moles } \text{O}_2} \][/tex]

2. Calculating the Moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] that React:
Using the given number of moles of [tex]\( \text{O}_2 \)[/tex] (12.0 moles) and the mole ratio from the balanced equation, we can find the moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] that will react:

[tex]\[ \text{Moles of } \text{C}_2\text{H}_2 = 12.0 \text{ mol O}_2 \times \frac{2 \text{ mol } \text{C}_2\text{H}_2}{5 \text{ mol } \text{O}_2} = 4.8 \text{ mol } \text{C}_2\text{H}_2 \][/tex]

3. Calculating the Volume of [tex]\( \text{C}_2\text{H}_2 \)[/tex] that Reacts:
At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. We use this fact to convert the moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] to liters:

[tex]\[ \text{Volume of } \text{C}_2\text{H}_2 = 4.8 \text{ mol } \text{C}_2\text{H}_2 \times 22.4 \text{ L/mol } \text{C}_2\text{H}_2 \][/tex]

4. Performing the Calculation:
[tex]\[ \text{Volume of } \text{C}_2\text{H}_2 = 4.8 \text{ mol } \text{C}_2\text{H}_2 \times 22.4 \text{ L/mol } = 107.52 \text{ L} \][/tex]

So, 107.52 liters of [tex]\( \text{C}_2\text{H}_2 \)[/tex] react with 12.0 moles of [tex]\( \text{O}_2 \)[/tex] at STP.
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