C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)

What volume, in L, of CO₂ gas is produced at STP when 2.00 mol C₂H₅OH reacts?

(Remember: 1 mole gas = 22.4 L at STP)

Volume (L) CO₂:



Answer :

To determine the volume of [tex]\( CO_2 \)[/tex] gas produced at STP when 2.00 moles of [tex]\( C_2H_5OH \)[/tex] react, follow these steps:

1. Identify the stoichiometric relationship:
The balanced chemical equation is:
[tex]\[ C_2H_5OH (l) + 3O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (g) \][/tex]
According to the equation, 1 mole of [tex]\( C_2H_5OH \)[/tex] produces 2 moles of [tex]\( CO_2 \)[/tex].

2. Calculate the moles of [tex]\( CO_2 \)[/tex]:
Given that 2.00 moles of [tex]\( C_2H_5OH \)[/tex] are reacting,
[tex]\[ \text{Moles of } CO_2 \text{ produced} = 2.00 \text{ moles of } C_2H_5OH \times \left(\frac{2 \text{ moles of } CO_2}{1 \text{ mole of } C_2H_5OH}\right) = 4.00 \text{ moles of } CO_2 \][/tex]

3. Convert the moles of [tex]\( CO_2 \)[/tex] to volume at STP:
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters.
[tex]\[ \text{Volume of } CO_2 \text{ at STP} = 4.00 \text{ moles of } CO_2 \times 22.4 \text{ liters/mole} = 89.6 \text{ liters} \][/tex]

4. State the final answer:
The volume of [tex]\( CO_2 \)[/tex] produced at STP when 2.00 moles of [tex]\( C_2H_5OH \)[/tex] react is [tex]\( 89.6 \)[/tex] liters.