Consider the following reaction equation C2H4(g) + 3O2(g) \(\to\) 2CO2(g) + 2H2O(l). The volume of oxygen at s.t.p that will be required to burn 14g of ethe...

Consider the following reaction equation
C₂H_4(g) + 3O_2(g) \(\to\) 2CO_2(g) + 2H₂O_(l).

The volume of oxygen at s.t.p that will be required to burn 14g of ethene is
[C₂H₄ = 28; Molar volume of gas at s.t.p = 22.4dm³

Answer Details

To solve this problem, we need to first balance the chemical equation: C₂H_4(g) + 3O_2(g) → 2CO_2(g) + 2H₂O_(l)
From the balanced equation, we can see that 1 mole of ethene reacts with 3 moles of oxygen gas to produce 2 moles of carbon dioxide gas and 2 moles of liquid water.
We can calculate the number of moles of ethene using the given mass and molar mass: n(C₂H₄) = m/M = 14/28 = 0.5 moles
Using the mole ratio from the balanced equation, we can calculate the number of moles of oxygen required: n(O₂) = 3n(C₂H₄) = 1.5 moles
Finally, we can use the ideal gas law to calculate the volume of oxygen at s.t.p: PV = nRT V = nRT/P = (1.5 mol)(8.31 J/mol K)(273 K)/(101,325 Pa) = 33.6 dm³
Therefore, the volume of oxygen at s.t.p that will be required to burn 14g of ethene is 33.6 dm³.
Answer: 33.6dm³.