combusion reichon, in the engine of your atooons water vapor (H:Ou), and great d

Question

combusion reichon, in the engine of your atooons water vapor (H:Ou), and great deal of energy the following Balance the equation for this reaction, and use dimensional analysis to answer questions: (INCLUDE UNITS! and do unit cancellation. Pay attention to significant figures) a. 3110 grams (I gallon) of cyclo-heptane (CHien) are chemically combined with oxygen (Oxg) in a this chemical reaction. Calculate the number of moles present in the 3110 grams of cyclo-heptane (C-Hi4D). 311 b. If the number of moles of CHi)calculated in (a) were burned in this combustion reaction, how many moles of oxygen (Onp) would be required to react with the cyclo-heptane? c. If the number of moles of C,Hi4u calculated in (a) were burned in this combustion reaction, how many moles and how many grams of carbon dioxide (COp) would be produced in the reaction? d. If the number of moles of C HiaD calculated in (a) were burned in this combustion reaction, how many moles and how many grams of water vapor (H2Oso) would be produced in the reaction? e. Use the number of moles (n) of carbon dioxide, calculated in part "c", and the Ideal Gas Law to calculate the volume of carbon dioxide gas (COag) produced in this reaction when the carbon dioxide gas is at a temperature of 1050.0°C and a pressure of 775 mmHg.

Explanation / Answer

Part a

Balanced equation:

2 C7H14 + 21 O2 = 14 CO2 + 14 H2O

Mass of C7H14 = 3110 g

Moles of C7H14 = mass/molecular weight

= 3110 g / 98.18

= 31.676 mol

Part b

Moles of O2 require

= 21 mol O2 x 31.676 mol C7H14 / 2 mol C7H14

= 332.598 mol O2

Part C

Moles of CO2 produced = 14 mol O2 x 31.676 mol C7H14 / 2 mol C7H14

= 221.732 mol CO2

Mass of CO2 produced = moles x molecular weight

= 221.732 mol x 44 g/mol

= 9756.208 g

Part d

Moles of H2O produced = 14 mol H2O x 31.676 mol C7H14 / 2 mol C7H14

= 221.732 mol H2O

Mass of H2O produced = moles x molecular weight

= 221.732 mol x 18 g/mol

= 3991.176 g

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