For the reaction
2Al+3CuSO4->Al2(SO4)3+3Cu
a maximum of ? moles of CO2 could be
formed from 6.5 mol of C6H6 and 4.56 mol of O2.
Am I dreaming, or what? Somebody needs some coffee to wake up.
No im just having some trouble... :(
I don't know that anyone would be surprised that you are having trouble converting benzene and O2 to Al2(SO4)3 and Cu.
To determine the maximum number of moles of CO2 that could be formed from the given amounts of C6H6 (benzene) and O2, we need to first balance the equation for the combustion of benzene.
The balanced equation for the combustion of benzene can be written as:
C6H6 + 15O2 → 6CO2 + 3H2O
From the balanced equation, we can see that 1 mole of benzene (C6H6) reacts with 15 moles of oxygen (O2) to produce 6 moles of carbon dioxide (CO2).
Now, let's calculate the moles of CO2 produced based on the given amounts of C6H6 and O2:
Given:
Moles of C6H6 = 6.5 mol
Moles of O2 = 4.56 mol
Using stoichiometry, we can determine the limiting reactant (the reactant that will be completely consumed) by comparing the ratio of moles between C6H6 and O2 in the balanced equation. The reactant with the smaller ratio is the limiting reactant.
From the balanced equation, we can see that the ratio of moles between C6H6 and O2 is 1:15. Therefore, 1 mole of C6H6 requires 15 moles of O2.
Moles of O2 required = Moles of C6H6 × (15 moles O2 / 1 mole C6H6)
= 6.5 mol × 15
= 97.5 mol
Since we have only 4.56 moles of O2, which is less than the required 97.5 moles, O2 is the limiting reactant.
Now we can use the stoichiometry from the balanced equation to determine the moles of CO2 produced from the limiting reactant (O2).
Moles of CO2 produced = Moles of O2 × (6 moles CO2 / 15 moles O2)
= 4.56 mol × (6/15)
≈ 1.824 mol
Therefore, a maximum of approximately 1.824 moles of CO2 could be formed from 6.5 mol of C6H6 and 4.56 mol of O2 in the given reaction.