Question

5.90

The automobile fuel called E85 consists of 85% ethanol and 15% gasoline. E85 can be used in so-called flex-fuel vehicles (FFVs), which can use gasoline, ethanol, or a mix as fuels. Assume that gasoline consists of a mixture of octanes (different isomers of C8H18), that the average heat of combustion of C8H18(l) is 5400 kJ/mol, and that gasoline has an average density of 0.70 g/mL. The density of ethanol is 0.79 g/mL.

1) Assume that the density and heat of combustion of E85 can be obtained by using 85% of the values for ethanol and 15% of the values for gasoline. How much energy could be released by the combustion of 3.5 L of E85?

Answer 1

Answer on the Question #63518, Chemistry / General chemistry

The automobile fuel called E85 consists of 85% ethanol and 15% gasoline. E85 can be used in so-called flex-fuel vehicles (FFVs), which can use gasoline, ethanol, or a mix as fuels. Assume that gasoline consists of a mixture of octanes (different isomers of C8H18), that the average heat of combustion of C8H18(I) is 5400 kJ/mol, and that gasoline has an average density of 0.70 g/mL. The density of ethanol is 0.79 g/mL.

1) Assume that the density and heat of combustion of E85 can be obtained by using 85% of the values for ethanol and 15% of the values for gasoline. How much energy could be released by the combustion of 3.5 L of E85?

Solution:

The main part of energy calculation is the aim to understand which volume takes 85% of ethanol and 15% of gasoline in 3.5 L:

$$V(\text{Ethanol}) = V \cdot 0.85 = 3.5 \cdot 0.85 = 2.975 \, \text{L}$$ $$V(\text{gasoline}) = V \cdot 0.15 = 3.5 \cdot 0.15 = 0.525 \, \text{L}$$

The heat of combustion is the ration between the produced energy and number of moles of the substance releases energy by the combustion to carbon dioxide and water:

$$\Delta_c H^0 = \frac{E}{n}$$

The mole number of each fuel can be rationalize by the following equation:

$$n = \frac{m}{M} = \frac{d \cdot V}{M}$$ $$n(\text{Ethanol}) = \frac{0.79 \, \text{g/mL} \cdot 2.975 \cdot 10^{-3} \, \text{mL}}{46 \, \text{g/mol}} = 5.1 \cdot 10^{-5} \, \text{mol}$$ $$n(\text{gasoline}) = \frac{0.70 \, \text{g/mL} \cdot 0.525 \cdot 10^{-3} \, \text{mL}}{114 \, \text{g/mol}} = 3.2 \cdot 10^{-6} \, \text{mol}$$

The energy what releases by the combustion of the E85:

$$E = E(\text{Ethanol}) + E(\text{gasoline})$$ $$E(\text{Ethanol}) = \Delta_c H^0(\text{Ethanol}) \cdot n(\text{Ethanol}) = 1370 \, \frac{\text{kJ}}{\text{mol}} \cdot 5.1 \cdot 10^{-5} \, \text{mol} = 69.9 \, \text{J}$$ $$E(\text{gasoline}) = \Delta_c H^0(\text{gasoline}) \cdot n(\text{gasoline}) = 5400 \, \frac{\text{kJ}}{\text{mol}} \cdot 3.2 \cdot 10^{-6} \, \text{mol} = 17.3 \, \text{J}$$ $$E = E(\text{Ethanol}) + E(\text{gasoline}) = 69.9 \, \text{J} + 17.3 \, \text{J} = 87.2 \, \text{J}$$

Answer: the energy releases by the E85 combustion equal to 87.2 J.

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