For the reaction
? C6H6 + ? O2 ↽⇀? CO2 + ? H2O a maximum of ? moles of CO2 could be formed from 9 mol of C6H6 and 2.81 mol of O2.
Answer in units of mol
This limiting reagent problem is done the same way as the one you posted above.
To find the maximum moles of CO2 that could be formed from the given amounts of C6H6 and O2, we need to use the balanced chemical equation for the combustion of benzene (C6H6):
C6H6 + O2 → CO2 + H2O
First, let's calculate the moles of C6H6 present in the reaction:
Given: 9 mol of C6H6
So, we have ? moles of C6H6 = 9 mol
Next, let's calculate the moles of O2 present in the reaction:
Given: 2.81 mol of O2
So, we have ? moles of O2 = 2.81 mol
To determine the maximum moles of CO2 that could be formed, we need to compare the ratios of moles of C6H6 and O2 to CO2 from the balanced equation.
From the balanced equation, we can see that the ratio of C6H6 to CO2 is 1:1, and the ratio of O2 to CO2 is 1:6. This means that for every 1 mole of C6H6, 1 mole of CO2 is formed, and for every 1 mole of O2, 6 moles of CO2 are formed.
So, let's calculate the maximum moles of CO2 formed by considering the limiting reactant (the reactant that is completely consumed) between C6H6 and O2:
For C6H6: From the given amount of 9 mol, the maximum moles of CO2 formed would be 9 mol.
For O2: From the given amount of 2.81 mol, the maximum moles of CO2 formed would be (2.81 mol) x (6 mol of CO2 / 1 mol of O2) = 16.86 mol of CO2.
Since the limiting reactant is O2 with a maximum of 16.86 moles of CO2 formed, the actual maximum moles of CO2 formed from 9 mol of C6H6 and 2.81 mol of O2 would be 16.86 mol.
Therefore, the maximum moles of CO2 formed is 16.86 mol.