Question 1 Report
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 ethene is
[C2H4 = 28; Molar volume of gas at s.t.p = 22.4dm3
To solve this problem, we need to first balance the chemical equation: C2H4(g) + 3O2(g) → 2CO2(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.