Answer:
24
Explanation:
First, find the limiting reagent by seeing how much reactant we'd have left if each of them was the limiting reagent and subtracting it by the initial.
Assume first that C6H6O is the limiting reagent. This means that all 4 moles of C6H6O will be used up in the reaction. Using stoichiometry, we know from the reaction that for every 1 mole of C6H6O, we use up 7 moles of O2 and produce 6 moles of CO2. (The numbers are from the coefficients of the chemicals in the reaction.)
So, we can now write out the first equation:
4 mol C6H6O | 7 mol O2 |
------------------------------------------------- = 28 mol O2
| 1 mol C6H6O |
The above equations says that if used up all 4 moles of C6H6O, we'd have to use up 28 mol of O2. We subtract this by the initial to see if the amount of O2 left is positive or negative.
35 mol O2 - 28 mol O2 = 7 mol O2
Since it is positive, we know that C6H6O is the limiting reagent.
For practice or to check work, you can also solve if O2 was the limiting reagent by confirming that you get a negative mole value at the end.
35 mol O2 | 1 mol C6H6O |
---------------------------------------------- = 5 mol C6H6O
| 7 mol O2 |
4 mol C6H6O - 5 mol C6H6O = -1 mol C6H6O
Since we can't have negative moles of reactant, C6H6O is the limiting reagent.
Next, we use stoichiometry to find the amount of CO2. From the reaction, we know that 1 mol of C6H6O consumed gives 6 mols of CO2. We position the ration so that the 'mol C6H6O' units cancel out and solve.
4 mol C6H6O * (6 mol CO2 / 1 mol C6H6O) = 24 mol CO2.
