In: Chemistry
1. Write a balanced equation for the complete combustion of each of the following:
a. 2,2-dimethylbutane
Express your answer as a chemical equation. Identify all of the phases in your answer.
b. cyclopentane
Express your answer as a chemical equation. Identify all of the phases in your answer.
2. Consider the compound ethylcyclopentane
a. Write the equation for the complete combustion of ethylcyclopentane.
Express your answer as a chemical equation.
b. Calculate the grams O2 required for the reaction of 35.0 g ethylcyclopentane.
Express your answer with the appropriate units.
c. How many liters of CO2 would be produced at STP from the reaction in part b?
Express your answer with the appropriate units.
Ans 1:
a)
Chemical formula of 2,2-dimethylbutane is C6H14.
Complete combustion of 2,2-dimethylbutane involves its reaction with oxygen to form water and carbon dioxide. Balanced equation of this combustion reaction can be shown as follows:
2C6H14(l) + 19O2(g) --------> 12CO2(g) + 14H2O(g)
b)
Chemical formula of cyclopentane is C5H10.
Complete combustion of cyclopentane involves its reaction with oxygen to form water and carbon dioxide. Balanced equation of this combustion reaction can be shown as follows:
2C5H10(l) + 15O2(g) --------> 10CO2(g) + 10H2O(g)
Ans2:
a)
Chemical formula of ethylcyclopentane is C7H14.
Complete combustion of ethylcyclopentane involves its reaction with oxygen to form water and carbon dioxide. Balanced equation of this combustion reaction can be shown as follows:
2C7H14(l) + 21O2(g) --------> 14CO2(g) + 14H2O(g)
b)
Balanced equation of combustion of ethylcyclopentane indicates that two moles of ethylcyclopentane requires 21 moles of oxygen gas. Molar mass of ethylcyclopentane is 98g/mol and that of oxygen is 32g/mol.
Number of moles of ethylcyclopentane in 35g can be calculated as follows:
Moles of ethylcyclopentane = (35 g) / (98g/mol)
Moles of ethylcyclopentane = (35 g) / (98g/mol)
Moles of ethylcyclopentane = 0.36 mol
Number of moles of oxygen required for combustion of 0.36moles of ethylcyclopentane can be calculated as follows:
Moles of oxygen = (21 moles of O2 / 2 mol of ethylcyclopetane) x 0.36 mol of ethylcyclopentane
Moles of oxygen = 3.78 mol
Thus, weight of 3.78 moles of O2 can be calculated as follows:
Weight of O2 = 3.78 mol x 32 g/mol
Weight of O2 = 121 g
Therefore, 121g of O2 is required for the complete combustion of 35.0 g ethylcyclopentane.
c)
Balanced equation of combustion of ethylcyclopentane indicates that two moles of ethylcyclopentane produces 14 moles of CO2. Since, 35g of ethylcyclopentane is 0.36 moles of ethylcyclopentane, therefore, moles of CO2 produced from 35g of ethylcyclopentane can be calculated as follows:
Moles of CO2 = (14 moles of CO2 / 2 mol of ethylcyclopetane) x 0.36 mol of ethylcyclopentane
Moles of CO2 = 2.52 mol
One mole of an ideal gas has a volume of 22.4 liters at STP. Thus, volume of 2.52 moles of CO2 can be calculated as follows:
Volume of CO2 = (2.52 x 22.4)L
Volume of CO2 = 56.4 L
Therefore, 56.4 liters of CO2 would be produced at STP from the reaction of 35.0 g ethylcyclopentane.