How many liters of oxygen is needed to burn 10g of polyethylene?

(-CH₂-CH₂-) n – polyethylene
(-CH₂-CH₂-) n can be written as n * C₂H₄
0.36 x
C₂H₄ + 3 / 2n * O₂ = * nCO₂ + nH₂O – the equation is the same as for alkenes.
1 mol 3/2 * 2 = 3 mol
or n (-CH₂-CH₂-) = n (O₂) / 3 or n (O₂) = 3 * n (-CH₂-CH₂-)
M (-CH₂-CH₂-) = 12 * 2 + 4 * 1 = 28 g / mol
n (-CH₂-CH₂-) = m / M = 10 g / 28 g / mol = 0.36 mol
3n (O₂) = 0.36 mol
n (O₂) = 0.36 * 3 mol = = 1.08 mol
V (O₂) = n * Vm = 1.08 mol * 22.4 L / mol = 24.2 L