Gasahol is a fuel containing ethanol(C2H6O) that burns in oxygen (O2)to give carbon dioxide and water. a)How many moles of O2 are needed to completely react with 4.0 moles of C2H60? b) If a car produces 88g of CO2,how many grams of O2 are used up in the reaction? c)If you add 125g of C2H6O to your gas, how many grams of CO2 and H2O can be produced?
a) To determine the number of moles of O2 required to react with 4.0 moles of C2H6O, we need to examine the balanced chemical equation for the reaction.
The balanced equation for the combustion of C2H6O is:
C2H6O + O2 → CO2 + H2O
According to the stoichiometry of the equation, the molar ratio of C2H6O to O2 is 1:1. This means that for every 1 mole of C2H6O, we need 1 mole of O2.
Therefore, if we have 4.0 moles of C2H6O, we would also need 4.0 moles of O2.
b) To determine the grams of O2 used up, we need to calculate the molar mass of CO2 and use the stoichiometry of the balanced equation.
The molar mass of CO2 is calculated as follows:
C: 1 atom × 12.01 g/mol = 12.01 g/mol
O: 2 atoms × 16.00 g/mol = 32.00 g/mol
Total molar mass of CO2: 12.01 g/mol + 32.00 g/mol = 44.01 g/mol
According to the balanced equation, the molar ratio of CO2 to O2 is 1:1. This means that for every 1 mole of CO2 produced, 1 mole of O2 is consumed.
Using the molar mass of CO2, we can calculate that 1 mole of CO2 is equal to 44.01 g.
If the car produces 88 g of CO2, we can set up a proportion to determine the grams of O2 used up:
88 g CO2 / 44.01 g/mol CO2 = x g O2 / 1 mol O2
Simplifying the equation, we find:
x = (88 g CO2 / 44.01 g/mol CO2) × 1 mol O2
x ≈ 2 mol O2
To convert the moles of O2 to grams, we can use the molar mass of O2, which is approximately 32.00 g/mol.
2 mol O2 × 32.00 g/mol O2 = 64 g O2
Therefore, approximately 64 g of O2 are consumed in the reaction.
c) To determine the grams of CO2 and H2O produced when 125 g of C2H6O is burned, we need to use the stoichiometry of the balanced equation.
First, we can calculate the moles of C2H6O in 125 g by using its molar mass:
C: 2 atoms × 12.01 g/mol = 24.02 g/mol
H: 6 atoms × 1.01 g/mol = 6.06 g/mol
O: 1 atom × 16.00 g/mol = 16.00 g/mol
Total molar mass of C2H6O: 24.02 g/mol + 6.06 g/mol + 16.00 g/mol = 46.08 g/mol
Using molar mass, we can calculate the moles of C2H6O:
125 g C2H6O / 46.08 g/mol C2H6O ≈ 2.71 mol C2H6O
According to the balanced equation, the stoichiometric molar ratio of C2H6O to CO2 is 1:1, and the molar ratio of C2H6O to H2O is 1:1 as well.
Therefore, 2.71 mol of C2H6O will produce approximately 2.71 mol of CO2 and 2.71 mol of H2O.
To calculate the grams of CO2 and H2O, we need to use their respective molar masses:
CO2: 44.01 g/mol
H2O: 18.02 g/mol
Calculating the grams of CO2 and H2O produced:
CO2: 2.71 mol CO2 × 44.01 g/mol CO2 ≈ 119.29 g CO2
H2O: 2.71 mol H2O × 18.02 g/mol H2O ≈ 48.86 g H2O
Therefore, approximately 119.29 g of CO2 and 48.86 g of H2O can be produced.