Answered

Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Get quick and reliable solutions to your questions from a community of experienced experts on our 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.
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.