Answered

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

How many moles of [tex]\(\text{H}_2\text{O}\)[/tex] form from [tex]\(13.5\,L\,\text{O}_2\)[/tex], assuming the reaction is at STP?

[tex]\[
\begin{array}{c|c|c}
13.5\,L\,\text{O}_2 & \dfrac{1\,mol\,\text{O}_2}{22.4\,L\,\text{O}_2} & \dfrac{2\,mol\,\text{H}_2\text{O}}{5\,mol\,\text{O}_2} \\
\hline
& & \\
\end{array}
\][/tex]

[tex]\[ ? \, mol\,\text{H}_2\text{O} \][/tex]



Answer :

GinnyAnswer
To determine how many moles of [tex]\(H_2O\)[/tex] (water) form from 13.5 liters of [tex]\(O_2\)[/tex] (oxygen) at standard temperature and pressure (STP), we need to follow the balanced chemical equation given:

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

The steps to solve this problem are as follows:

1. Determine the molar volume at STP:
At STP (standard temperature and pressure), one mole of an ideal gas occupies 22.4 liters.

2. Calculate the moles of [tex]\(O_2\)[/tex]:
Given that you have 13.5 liters of [tex]\(O_2\)[/tex], we use the molar volume at STP to find the number of moles of [tex]\(O_2\)[/tex]:

[tex]\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}} = \frac{13.5 \text{ L}}{22.4 \text{ L/mol}} \approx 0.6027 \text{ mol} \][/tex]

3. Use the stoichiometry of the balanced equation:
From the balanced chemical equation, the ratio of moles of [tex]\(O_2\)[/tex] to moles of [tex]\(H_2O\)[/tex] is 5:2. This means that for every 5 moles of [tex]\(O_2\)[/tex], 2 moles of [tex]\(H_2O\)[/tex] are produced. To find the moles of [tex]\(H_2O\)[/tex] formed from the calculated moles of [tex]\(O_2\)[/tex], we set up the ratio:

[tex]\[ \text{Moles of } H_2O = \left(\text{Moles of } O_2\right) \times \frac{2 \text{ mol } H_2O}{5 \text{ mol } O_2} \][/tex]

4. Calculate the moles of [tex]\(H_2O\)[/tex]:
Plug in the moles of [tex]\(O_2\)[/tex] to find the moles of [tex]\(H_2O\)[/tex]:

[tex]\[ \text{Moles of } H_2O = 0.6027 \text{ mol} \times \frac{2}{5} \approx 0.2411 \text{ mol} \][/tex]

Therefore, the number of moles of [tex]\(H_2O\)[/tex] formed from 13.5 liters of [tex]\(O_2\)[/tex] at STP is approximately 0.2411 moles.