Question

n the following reaction, how many liters of oxygen will react with 270 liters of ethene (C2H4) at STP? C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(g) The volume of one mole of gas at STP is 22.4 liters.

Answer 1

C₂H₄ + 3O₂ = 2CO₂ + 2H₂O

V(C₂H₄) = 270 L

V(O₂)=3V(C₂H₄)

V(O₂)=3*270=810 L

Answer 2

Answer: 810 L

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given volume}}{\text{Molar volume}}$

For $C_2H_4$

Given volume= 270 L

Molar volume of  $C_2H_4$ = 22.4 L

Putting values in above equation, we get:

$\text{Moles of}C_2H_4 =\frac{270}{22.4}=12.05mol$

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)$

1 mole of $C_2H_4$ reacts with 3 moles of $O_2$

Thus 12.05 moles $C_2H_4$ reacts with =$\frac{3}{1}\times 12.05=36.15$ moles of $O_2$

Volume of $O_2=moles\times {\text {molar volume}}=36.15\times 22.4=810L$

Thus 810 L of oxygen will react with 270 L of ethene.