offered 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 countless liters of carbon dioxide will be produced under typical conditions? The thickness of cyclohexane is 0.75 g/ml.
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This is a stoichiometry question. Most stoichiometry calculations use one or more of the adhering to mole conversion factors:

1 mole (mol) = molar mass of the substance in grams (molar mass have to be calculated because that each substance)1 mole (mol) = 22.4 liters (L) for appropriate gases at STP 1...

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This is a stoichiometry question. Many stoichiometry calculations use one or much more of the adhering to mole counter factors:

1 mole (mol) = molar massive of the problem in grams (molar mass must be calculated because that each substance)1 mole (mol) = 22.4 liters (L) for ideal gases in ~ STP1 mole (mol) = 6.02 x ~10^23 particles (molecules or atoms)

Step 1: convert both offered substances to moles. The reason we have to do this is that us will need to use mole coefficients afterwards to determine the limiting reactant and also to execute the stoichiometry calculation.

Convert 86.7 l ~O_2 to moles: We can transform the 86.7 l of ~O_2 to moles by utilizing the second mole switch factor presented above:

(86.7 L)(1 mol/22.4 L) = 3.87 mol ~O_2` `

Convert 48 mL ~C_6H_12 to moles: Since no one of the mole conversion components shown over contains mL, us must first convert "mL" to grams. Us can convert from mL to grams by using the thickness given in the problem as a switch factor. Climate we have the right to use the mole conversion element containing grams (molar mass) shown above to convert to moles.

(48 mL)(0.75 g/1 mL) = 36 g ~C_6H_12

In stimulate to usage the conversion factor for moles and also molar mass, we should 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 element that has the molar mass to transform to moles:

(36 g)(1 mol/84.162 g) = 0.43 mol ~C_6H_12

Step 2: determine the limiting reactant: Because you were provided two reactants, you require to determine if among them is a limiting reactant. The limiting reactant is the reactant that runs out first. The limiting reactant will certainly be used as the beginning substance because that the stoichiometry calculate in step 3.

To determine the limiting reactant, division the moles of every reactant by the reactant"s coefficient. The coefficients are discovered in the balanced equation.

Reactant ~O_2: 3.87 mol/9 = 0.43Reactant ~C_6H_12: 0.43 mol/1 = 0.43

Compare the 2 answers. The reactant through the smallest answer is the limiting reactant. Because both answers space the same, the reactants will certainly run the end at the same time and also neither reactant boundaries the reaction. Since of this, we can use either of the reaction to begin the stoichiometry calculation. You should acquire the same answer nevertheless of i m sorry reactant you usage to begin the stoichiometry calculation.

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Step 3: execute the stoichiometry calculation: The basic pattern of the stoichiometry calculation is:

(moles that reactant) x (coefficient ratio: product/reactant) x (mole counter factor: 1 mol = 22.4 L)

Since neither reactant is limiting, let"s execute the stoichiometry difficulty using both that the reactants as the starting substance. We should get the same amount the ~CO_2 created 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) =