Given the complete combustion of cyclohexane (C6H12 + 9O2 ---> 6CO2 + 6H2O), if 48 ml of cyclohexane are reacted with 86.7 liters of oxygen at STP, how many liters of carbon dioxide will be...
Given the complete combustion of cyclohexane (C6H12 + 9O2 ---> 6CO2 + 6H2O), if 48 ml of cyclohexane are reacted with 86.7 liters of oxygen at STP, how many liters of carbon dioxide will be produced under standard conditions? The density of cyclohexane is 0.75 g/ml.
This is a stoichiometry question. Most stoichiometry calculations use one or more of the following mole conversion factors:
- 1 mole (mol) = molar mass of the substance in grams (molar mass must be calculated for each substance)
- 1 mole (mol) = 22.4 liters (L) for ideal gases at STP
- 1 mole (mol) = 6.02 x ~10^23 particles (molecules or atoms)
Step 1: Convert both given substances to moles. The reason we need to do this is that we will need to use mole coefficients later on to determine the limiting reactant and to perform the stoichiometry calculation.
- Convert 86.7 L ~O_2 to moles: We can convert the 86.7 L of ~O_2 to moles by using the second mole conversion factor shown above:
(86.7 L)(1 mol/22.4 L) = 3.87 mol ~O_2` `
- Convert 48 mL ~C_6H_12 to moles: Since none of the mole conversion factors shown above contains mL, we must first convert "mL" to grams. We can convert from mL to grams by using the density given in the problem as a conversion factor. Then we can use the mole conversion factor containing grams (molar mass) shown above to convert to moles.
(48 mL)(0.75 g/1 mL) = 36 g ~C_6H_12
In order to use the conversion factor for moles and molar mass, we must calculate the molar mass of ~C_6H_12:
Molar mass = (6 x 12.011) + (12 x 1.008) = 84.162 g ~C_6H_12
Now we can use the mole conversion factor that contains the molar mass to convert to moles:
(36 g)(1 mol/84.162 g) = 0.43 mol ~C_6H_12
Step 2: Determine the limiting reactant: Because you were given two reactants, you need to determine if one of them is a limiting reactant. The limiting reactant is the reactant that runs out first. The limiting reactant will be used as the starting substance for the stoichiometry calculation in Step 3.
To determine the limiting reactant, divide the moles of each reactant by the reactant's coefficient. The coefficients are found in the balanced equation.
- Reactant ~O_2: 3.87 mol/9 = 0.43
- Reactant ~C_6H_12: 0.43 mol/1 = 0.43
Compare the two answers. The reactant with the smallest answer is the limiting reactant. Since both answers are the same, the reactants will run out at the same time and neither reactant limits the reaction. Because of this, we can use either of the reactants to begin the stoichiometry calculation. You should get the same answer regardless of which reactant you use to begin the stoichiometry calculation.
Step 3: Perform the stoichiometry calculation: The general pattern of the stoichiometry calculation is:
(moles of reactant) x (coefficient ratio: product/reactant) x (mole conversion factor: 1 mol = 22.4 L)
Since neither reactant is limiting, let's perform the stoichiometry problem using both of the reactants as the starting substance. We should get the same amount of ~CO_2 produced either way.
- Stoichiometry calculation starting with moles of ~O_2:
(3.87 mol ~O_2)(6 mol ~CO_2/9 mol ~O_2)(22.4 L/1 mol) =
57.8 L ~CO_2
- Stoichiometry calculation starting with moles of ~C_6H_12:
(0.43 mol ~C_6H_12)(6 mol ~CO_2/1 mol ~C_6H_12)(22.4 L/ 1 mol) =
57.8 L CO_2