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How do I calculate what mass of CO2 can be by mixing 2.34g of C2H5OH with 5.6L oxygen at 78°F

The reaction occurs at 900 mmHg and the formula for the reaction is: C2H5OH+O2===> CO2+H2O

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LaRita W. | Master's in Chemistry - AP Chemistry Teacher - Math and Science TutorMaster's in Chemistry - AP Chemistry Tea...
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To determine the mass of CO2 product that can be formed, we'll have to do some stoichiometry. But before we get there, we'll have to figure out which of our reactants (O2 or C2H5OH) is the limiting reactant based on the number of moles present and the mole ratio at which they come together.
 
From the conditions given, we can use the ideal gas law equation to determine the moles of O2 gas present:
PV = nRT
(1.18 atm)(5.6L) = n(http://0.08206L.atm/mol.K)(298 K)
n= 0.270 molesO2
*notice, I converted the pressure given into atm and the temperature given into Kelvin first so that thse units matched the units in our ideal gas constant, R.
 
From the given conditions, we can also find the moles of the other reactant C2H5OH:
2.34g C2H5OH x (1mole/46g) = 0.0509 moles C2H5OH
 
According to the balanced equation:
C2H5OH + 3O2===> 2CO2 + 3H2O
these reactants come together in a 1:3 ratio. Therefore, C2H5OH is limiting, so we will use that compound to predict how much product can be formed:
 
0.0509 moles C2H5OH x (2 moles CO2/1mole C2H5OH) x (44g CO2/1 mole CO2) = 4.48 g CO2
 
J.R. S. | Ph.D. in Biochemistry--University Professor--Chemistry TutorPh.D. in Biochemistry--University Profes...
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C2H5OH + 3O2===> 2CO2 + 3H2O .. balanced equation
moles C2H5OH present = 2.34 g x 1 mole/46 g = 0.0509 moles
moles O2 present = (900 mm)(5.6L)/(62.36 Lmm/Kmole)(351K) = 0.230 moles O2
Limiting reactant is C2H5OH
moles CO2 formed = 0.0509 moles C2H5OH x 2 moles CO2/mole C2H5OH = 0.1018 moles CO2
mass CO2 = 0.1018 moles x 44 g/mole = 4.48 g = 4.5 g (2 significant figures)