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Given the reaction:

[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]

Calculate the number of liters of [tex]\(\text{C}_2 \text{H}_2\)[/tex] that react with 12.0 mol [tex]\(\text{O}_2\)[/tex], assuming the reaction is at STP.

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

[tex]\[
\begin{array}{c|c|c}
12.0 \, \text{mol} \, \text{O}_2 & 2 \, \text{mol} \, \text{C}_2 \text{H}_2 & \text{[?]} \\
\hline
& 5 \, \text{mol} \, \text{O}_2 & \text{[?]}
\end{array}
\][/tex]

Ratio 1:
[tex]\[
\frac{1 \, \text{mol} \, \text{C}_2 \text{H}_2}{26.04 \, \text{g} \, \text{C}_2 \text{H}_2}
\][/tex]

Ratio 2:
[tex]\[
\frac{26.04 \, \text{g} \, \text{C}_2 \text{H}_2}{1 \, \text{mol} \, \text{C}_2 \text{H}_2}
\][/tex]

Ratio 3:
[tex]\[
\frac{1 \, \text{mol} \, \text{C}_2 \text{H}_2}{22.4 \, \text{L} \, \text{C}_2 \text{H}_2}
\][/tex]



Answer :

GinnyAnswer
To solve the problem, we need to determine how many liters of [tex]\( \text{C}_2\text{H}_2 \)[/tex] will react with [tex]\( 12.0 \)[/tex] moles of [tex]\( \text{O}_2 \)[/tex] at STP (Standard Temperature and Pressure). We will go through the following steps:

1. Determine the molar ratio of [tex]\( \text{C}_2\text{H}_2 \)[/tex] to [tex]\( \text{O}_2 \)[/tex] from the balanced equation:
The balanced chemical equation is:
[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]
From this equation, we observe the molar ratio of [tex]\( \text{C}_2\text{H}_2 \)[/tex] to [tex]\( \text{O}_2 \)[/tex]:
[tex]\[ \frac{2 \text{ moles of } \text{C}_2\text{H}_2}{5 \text{ moles of } \text{O}_2} \][/tex]

2. Calculate the moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] required for [tex]\( 12.0 \)[/tex] moles of [tex]\( \text{O}_2 \)[/tex]:
[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. Calculate the volume of [tex]\( \text{C}_2\text{H}_2 \)[/tex] at STP:
At STP, one mole of any gas occupies [tex]\( 22.4 \)[/tex] liters. Therefore, the volume of [tex]\( 4.8 \)[/tex] moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] can be calculated as:
[tex]\[ \text{Volume of } \text{C}_2\text{H}_2 = 4.8 \text{ mol } \times 22.4 \text{ L/mol } = 107.52 \text{ L} \][/tex]

Summary:
After the calculations, it is found that [tex]\( 4.8 \)[/tex] moles of [tex]\( \text{C}_2\text{H}_2 \)[/tex] are required, and their volume at STP is [tex]\( 107.52 \)[/tex] liters. Therefore, [tex]\( 12.0 \)[/tex] moles of [tex]\( \text{O}_2 \)[/tex] will react with [tex]\( 107.52 \)[/tex] liters of [tex]\( \text{C}_2\text{H}_2 \)[/tex] at STP.

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