If 1.5 mol of each of the following compounds is completely combusted in oxygen, which one will produce the largest number of moles of H2O and why?
incomplete question.
To determine which compound will produce the largest number of moles of H2O when combusted, we need to analyze the stoichiometry of each compound.
The first step is to write the balanced chemical equations for the combustion of each compound. The general combustion formula is:
Compound + O2 → H2O + CO2
Compound A: C6H6 (Benzene)
C6H6 + 15O2 → 6CO2 + 3H2O
Compound B: C2H5OH (Ethanol)
C2H5OH + 3O2 → 2CO2 + 3H2O
Compound C: CH4 (Methane)
CH4 + 2O2 → CO2 + 2H2O
Next, we need to examine the stoichiometric ratios in each equation. The ratios indicate the number of moles of H2O produced per mole of the compound combusted.
For Compound A (Benzene):
The equation shows that for every 1 mole of C6H6 combusted, 3 moles of H2O are produced.
For Compound B (Ethanol):
The equation shows that for every 1 mole of C2H5OH combusted, 3 moles of H2O are produced.
For Compound C (Methane):
The equation shows that for every 1 mole of CH4 combusted, 2 moles of H2O are produced.
Now, since we have 1.5 moles of each compound, we can calculate the number of moles of H2O produced for each compound.
For Compound A (Benzene):
1.5 moles of C6H6 x 3 moles of H2O/mole of C6H6 = 4.5 moles of H2O
For Compound B (Ethanol):
1.5 moles of C2H5OH x 3 moles of H2O/mole of C2H5OH = 4.5 moles of H2O
For Compound C (Methane):
1.5 moles of CH4 x 2 moles of H2O/mole of CH4 = 3 moles of H2O
Given the calculations, we see that both Compound A (Benzene) and Compound B (Ethanol) produce the largest number of moles of H2O, which is 4.5 moles. This is more than the 3 moles produced by Compound C (Methane).
Therefore, both Compound A (Benzene) and Compound B (Ethanol) will produce the largest number of moles of H2O when completely combusted in oxygen.