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Question

Balance the reaction for the combustion of pentane below and enter the coefficient for oxygen.

 

 

C₅H₁₂ (l) + O₂ (g) → CO₂ (g) + H₂O (g)      ΔH_rxn = -3682 kJ/mol_rxn 

Using the previous equation that you balanced.

C₅H₁₂ (l) + O₂ (g) → CO₂ (g) + H₂O (g)      ΔH_rxn = -3682 kJ/mol_rxn 

how many moles of carbon dioxide (CO₂) can be produced from a 43.865 g sample of pentane , C₅H₁₂   ? Do not enter units after your answer.

Previously, you found the theoretical yield of CO_{2 .} If 26.0 L of carbon dioxide were recovered at 23.5 º C and 1.5 atm, what was the actual yield in grams? Do not enter units after your answer.

Explain the difference between the theoretical yield and actual yield. Within your explanation, explain where each of these values comes from and why these two values are not the same.

Answer 1

For the first part, we need to balance the chemical equation:

First, I'll balance the carbon atoms. We have 5 in the reactants and 1 in the products, so I add a coefficient of 5 to the carbon dioxide gas in the products.

Next, I'll balance the hydrogen atoms, and since there are six times as many in the reactants than in the products, I'll add a coefficient of 6 to the water vapor in the products.

Finally, I just need to balance the oxygen atoms. Since I added a coefficient of 6 to the water vapor and 5 to the carbon dioxide gas, I now have 16 oxygens in the products. So, to balance the reactants, I'll add a coefficient of 8 to the oxygen gas.

The next part of the question asks "how many moles of carbon dioxide (CO₂) can be produced from a

43.865 g sample of pentane , C₅H₁₂ ? "

I can answer this question using the mole ratios I determined by balancing the equation above. To begin, I'll convert grams of pentane into moles of pentane using pentane's molar mass. Pentane has a molar mass of 72.055 g/mol.

43.865 g C₅H₁₂  x [1 mol / 72.055 g] = 0.6088 mol C₅H₁₂ 

Now that pentane is in moles, we can set up a mol ratio to determine how many mols of carbon dioxide gas will be produced from 0.6088 moles of pentane (which is the same as 43.865 g of pentane). Using the balanced equation, 1 mol of pentane yields 5 moles of carbon dioxide gas.

0.6088 mol C₅H₁₂  x [5 mol CO₂ / 1 mol C₅H₁₂]  = 3.044 mol CO₂

The next step is to convert moles of CO₂ back into grams, and again, I'll use the molar mass of carbon dioxide to do so. The molar mass of carbon dioxide is 44.009 g/mol.

3.044 mol CO₂ x [44.009 g / 1 mol] = 133.96 g CO₂

So, the theoretical yield of carbon dioxide gas from the combustion of 43.865 grams of pentane is 133.96 grams.

The next question wants to know the actual yield of carbon dioxide in grams under conditions of 23.5 º C and 1.5 atm, where 26.0 L of CO₂ has been produced. Because we are give units of temperature, volume, and pressure, I will use the ideal gas law equation , PV = nRT, to calculate moles and then convert those moles to grams.

P = 1.5 atm

V = 26 L

T = 23.5 º C = 296.5 K

R = 0.08205 L·atm/mol·K

n = ? mol

Plugging the values into the equation, I get:

1.5 atm (26.0 L) = n (0.08205 L·atm/mol·K) (296.5 K)

n = 1.603 mol

I calculated the moles of carbon produced, and now I need to convert this value to grams using the molar mass of carbon dioxide, which we said earlier was 44.009 g/mol.

1.603 mol CO₂ x [44.009 g / 1 mol] = 70.55 g CO₂

This would be the actual yield of the chemical reaction. Theoretical yield is going to be the maximum amount of product we could produce under perfect conditions. Of course, when performing these reactions, the actual yield will nearly always fall under the theoretical yield because it is incredibly difficult to create perfect conditions. On top of that, there are shortcomings in the mechanisms for recovering the products, reactions which are incredibly slow, and the formation of side products which all detract from your yield.

Answer 2

A. 1 C₅H₁₂ (l) + 8 O₂ (g) → 5 CO₂ (g) + 6 H₂O (g)

B. 43.865g C₅H₁₂ [1 mol C₅H₁₂/72g C₅H₁₂] [5 mol CO₂/1 mol C₅H₁₂] ~ 3.0 mol CO₂ (Theoretical Yield)

3 mol CO₂ [44g CO₂/1 mol CO₂] = 132g CO₂ (Theoretical Yield in grams)

C. 26L CO₂ [44g CO₂/1mol CO₂] [1mol CO₂/22.4L CO₂] = 51g CO₂ (Actual Yield in grams @ STP)

STP: 273.15 K and 1 atm (Volume @ STP is 22.4L per mol)

Will F. demonstrates use of Ideal Gas Law to determine actual yield at 23.5 centigrade and 1.5atm. You can determine the R value by inputting the STP values into the Ideal Gas Law equation and solving for R:

PV = nRT

R = PV/nT = (1atm)(22.4L)/(1mol)(273.15K) = 0.0820 atm L/mol K

Now that you know R, you can arrange the equation to solve for n and plug in your values:

T = 23.5 centigrade = 296.65 K

n = PV/RT

n = (1.5atm)(26L)/(0.0820 atm L/mol K)(296.65 K) = 1.60 mol

Solve for grams CO₂:

1.60 mol CO₂ [44g CO₂/1 mol CO2] = 70.4g CO₂ (Actual Yield under described conditions)

D. 51g CO₂ [1 mol CO₂/44g CO₂] = 1.2 mol CO₂ (Actual Yield @ STP)

(Theoretical - actual)/Theoretical x 100 = % Error

(132 - 51)/132 x 100 = 61% (Difference between theoretical and values at STP)

(132 - 70.4)/132 x 100 = 47% (Difference between theoretical and values described in problem)

Theoretical values provide information regarding ideal conditions. Actual values are a reflection of experimental technique, experimental conditions, environment, etc.