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Gas Stoichiometry at STP

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

You want to calculate the number of liters of [tex]\[ C_2H_2 \][/tex] that react with 12.0 mol [tex]\[ O_2 \][/tex], assuming the reaction is at STP.

Use the dimensional analysis below to set up the two-step calculation:

[tex]\[
\begin{array}{c|c|c}
12.0 \text{ mol } O_2 & \rightarrow & \text{ ? L } C_2H_2 \\
\hline
\text{moles ratio} & 2 \text{ mol } C_2H_2 & \text{ ? L } \\
\hline
& 5 \text{ mol } O_2 &
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
Let's solve the given problem step-by-step.

### Step 1: Interpret the Reaction
The balanced chemical equation for the reaction is:
[tex]\[2 \, \text{C}_2\text{H}_2 + 5 \, \text{O}_2 \rightarrow 4 \, \text{CO}_2 + 2 \, \text{H}_2\text{O}\][/tex]

### Step 2: Identify the Problem
You're asked to calculate the number of liters of [tex]\( \text{C}_2\text{H}_2 \)[/tex] (acetylene) that react with 12.0 mol [tex]\( \text{O}_2 \)[/tex] (oxygen), under standard temperature and pressure (STP).

### Step 3: Determine the Mole Ratio
The reaction gives us the mole ratio between acetylene and oxygen:
[tex]\[2 \, \text{mol} \, \text{C}_2\text{H}_2 : 5 \, \text{mol} \, \text{O}_2\][/tex]

### Step 4: Calculate the Moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] Reacting with [tex]\( \text{O}_2 \)[/tex]
Using the mole ratio, we find the moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] that react with 12.0 mol [tex]\( \text{O}_2 \)[/tex]:
[tex]\[ \text{Moles of} \, \text{C}_2\text{H}_2 = \left(\frac{2 \, \text{mol} \, \text{C}_2\text{H}_2}{5 \, \text{mol} \, \text{O}_2}\right) \times 12.0 \, \text{mol} \, \text{O}_2 \][/tex]

From this, we calculate:
[tex]\[ \text{Moles of} \, \text{C}_2\text{H}_2 = \frac{2}{5} \times 12.0 \approx 4.8 \, \text{mol} \][/tex]

### Step 5: Convert Moles to Liters at STP
Under standard temperature and pressure (STP), 1 mole of a gas occupies 22.4 liters.

Thus, the volume of [tex]\( \text{C}_2\text{H}_2 \)[/tex] in liters can be found by multiplying the moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] by the volume occupied per mole:
[tex]\[ \text{Volume of} \, \text{C}_2\text{H}_2 = 4.8 \, \text{mol} \times 22.4 \, \text{L/mol} \][/tex]

Performing this multiplication:
[tex]\[ \text{Volume of} \, \text{C}_2\text{H}_2 \approx 107.52 \, \text{liters} \][/tex]

### Conclusion
Hence, 4.8 moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] react with 12.0 mol [tex]\( \text{O}_2 \)[/tex], and the volume of [tex]\( \text{C}_2\text{H}_2 \)[/tex] required is approximately 107.52 liters at STP.
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