Respuesta :
C₂H₄ + 3O₂ = 2CO₂ + 2H₂O
V(C₂H₄) = 270 L
V(O₂)=3V(C₂H₄)
V(O₂)=3*270=810 L
V(C₂H₄) = 270 L
V(O₂)=3V(C₂H₄)
V(O₂)=3*270=810 L
Answer: 810 L
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given volume}}{\text{Molar volume}}[/tex]
For [tex]C_2H_4[/tex]
Given volume= 270 L
Molar volume of [tex]C_2H_4[/tex] = 22.4 L
Putting values in above equation, we get:
[tex]\text{Moles of}C_2H_4 =\frac{270}{22.4}=12.05mol[/tex]
According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
[tex]C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)[/tex]
1 mole of [tex]C_2H_4[/tex] reacts with 3 moles of [tex]O_2[/tex]
Thus 12.05 moles [tex]C_2H_4[/tex] reacts with =[tex]\frac{3}{1}\times 12.05=36.15[/tex] moles of [tex]O_2[/tex]
Volume of [tex]O_2=moles\times {\text {molar volume}}=36.15\times 22.4=810L[/tex]
Thus 810 L of oxygen will react with 270 L of ethene.
