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Gas Stoichiometry at STP

The balanced chemical equation is:
[tex]\[ 2 C_2H_2(g) + 5 O_2(g) \rightarrow 4 CO_2(g) + 2 H_2O(g) \][/tex]

You want to calculate the number of moles of [tex]\( H_2O \)[/tex] that form from [tex]\( 13.5 \, L \, O_2 \)[/tex], assuming the reaction is at STP.

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

[tex]\[
\begin{array}{c|c|c}
13.5 \, L \, O_2 & \rightarrow & \text{moles of } O_2 \\
& \rightarrow & \text{moles of } H_2O
\end{array}
\][/tex]

Ratios:

1. [tex]\(\text{Ratio 1}\)[/tex]: [tex]\( \frac{\text{Moles of } O_2}{22.4 \, L} \)[/tex] (molar volume at STP)
2. [tex]\(\text{Ratio 2}\)[/tex]: [tex]\( \frac{\text{Moles of } H_2O}{\text{Moles of } O_2} \)[/tex] (from the balanced equation)



Answer :

GinnyAnswer
Let's solve this problem step-by-step using dimensional analysis as requested.

1. Step 1: Calculate the number of moles of [tex]\( O_2 \)[/tex] given: We know that at STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters.

[tex]\[ \text{Given volume of oxygen} = 13.5 \text{ L} \][/tex]

We use the molar volume to convert this to moles of [tex]\( O_2 \)[/tex]:

[tex]\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}} = \frac{13.5 \text{ L}}{22.4 \text{ L/mol}} = 0.6026785714285715 \text{ mol} \][/tex]

2. Step 2: Use the stoichiometric relationship to find the moles of [tex]\( H_2O \)[/tex] produced: The balanced chemical equation shows the stoichiometric relationship between [tex]\( O_2 \)[/tex] and [tex]\( H_2O \)[/tex]:

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

From the equation, 5 moles of [tex]\( O_2 \)[/tex] produce 2 moles of [tex]\( H_2O \)[/tex]. Therefore, the ratio between [tex]\( O_2 \)[/tex] and [tex]\( H_2O \)[/tex] is:

[tex]\[ \frac{\text{Moles of } H_2O}{\text{Moles of } O_2} = \frac{2}{5} \][/tex]

Now, using the moles of [tex]\( O_2 \)[/tex] calculated in Step 1:

[tex]\[ \text{Moles of } H_2O = \frac{2}{5} \times \text{Moles of } O_2 = \frac{2}{5} \times 0.6026785714285715 \approx 0.2410714285714286 \text{ mol} \][/tex]

Thus, the number of moles of [tex]\( H_2O \)[/tex] formed from 13.5 liters of [tex]\( O_2 \)[/tex] at STP is approximately 0.2410714285714286 moles.

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