Answered

Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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)

Sagot :

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.
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.