Question

1. What is the mole percent of ethanol (C₂H₅OH), which consists of 71.0 g of ethanol for every 10.0 g of water present?

a. 71.0%

b. 73.5%

c. 26.5%

d. 22.1%

e. 87.7%

2.  Ethanol, C₂H₆O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is a renewable resource and ethanol-blended fuels, like gasohol, appear to burn more efficiently in combustion engines. The heat of combustion of ethanol is 326.7 kcal/mol.

The heat of combustion of heptane, C₇H₁₆, is 1.151×10³ kcal/mol. How much energy is released during the complete combustion of 348 grams of heptane?

Assuming the same efficiency, would 348 grams of ethanol provide more, less, or the same amount of energy as 348 grams of heptane? More, less, or the same amount? Please type the answers! :)

Answer 1

1)

Molar mass of C2H5OH = 2*MM(C) + 6*MM(H) + 1*MM(O)

= 2*12.01 + 6*1.008 + 1*16.0

= 46.068 g/mol

Molar mass of H2O = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

n(C2H5OH) = mass/molar mass

= 71.0/46.068

= 1.5412

n(H2O) = mass/molar mass

= 10.0/18.016

= 0.5551

n(C2H5OH),n1 = 1.5412 mol

n(H2O),n2 = 0.5551 mol

Total number of mol = n1+n2

= 1.5412 + 0.5551

= 2.0963 mol

mole % of ethanol = n1*10-/total mol

= 1.5412*100/2.0963

= 73.5 %

Answer: b

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