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2 C[tex]\(_2\)[/tex]H[tex]\(_2\)[/tex](g) + 5 O[tex]\(_2\)[/tex](g) → 4 CO[tex]\(_2\)[/tex](g) + 2 H[tex]\(_2\)[/tex]O(g)

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

Given:
- 13.5 L O[tex]\(_2\)[/tex]
- 1 mol O[tex]\(_2\)[/tex] = 22.4 L O[tex]\(_2\)[/tex]
- 5 mol O[tex]\(_2\)[/tex] produces 2 mol H[tex]\(_2\)[/tex]O

Calculate the moles of H[tex]\(_2\)[/tex]O:

[tex]\[
\begin{aligned}
\text{Moles of O\(_2\)} &= \frac{13.5 \, \text{L O\(_2\)}}{22.4 \, \text{L O\(_2\) per mol}} \\
\text{Moles of H\(_2\)O} &= \frac{2 \, \text{mol H\(_2\)O}}{5 \, \text{mol O\(_2\)}} \times \text{Moles of O\(_2\)} \\
&= \frac{2}{5} \times \frac{13.5}{22.4} \\
&= [?] \, \text{mol H\(_2\)O}
\end{aligned}
\][/tex]

[tex]\[
\text{Moles of H\(_2\)O} = [?] \, \text{mol H\(_2\)O}
\][/tex]



Answer :

GinnyAnswer
To determine how many moles of [tex]\( H_2O \)[/tex] form from 13.5 L of [tex]\( O_2 \)[/tex] at Standard Temperature and Pressure (STP), we can follow this step-by-step procedure:

1. Determine the Moles of [tex]\( O_2 \)[/tex]:
- At STP, 1 mole of an ideal gas occupies 22.4 liters.
- Given volume of [tex]\( O_2 \)[/tex] = 13.5 L.
- We use the formula:
[tex]\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}} \][/tex]
Substituting the given values:
[tex]\[ \text{Moles of } O_2 = \frac{13.5 \text{ L}}{22.4 \text{ L/mol}} \approx 0.6027 \text{ moles} \][/tex]

2. Use the Stoichiometric Ratio to Find Moles of [tex]\( H_2O \)[/tex]:
- According to the balanced chemical equation:
[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 number of moles of [tex]\( H_2O \)[/tex] produced can be calculated using the proportionality:
[tex]\[ \text{Moles of } H_2O = \left( \frac{2 \text{ moles } H_2O}{5 \text{ moles } O_2} \right) \times \text{Moles of } O_2 \][/tex]
Substituting the known value:
[tex]\[ \text{Moles of } H_2O = \left( \frac{2}{5} \right) \times 0.6027 \approx 0.2411 \text{ moles} \][/tex]

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

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