C2H5OH + 5O2 --> 3CO2 + 4H2O?
Given 212 g of C2H5OH and 115g of O2
a) calculate the mass of the water vapor that will be produced
b)Identify the limiting reactant
c)calculate the percentage yield if 68.0 g of water vapor is actually produced
e)calculate the mass of unused reactant.
Comments
The equation above^ seems unbalanced
The # Oxygen on reactants is 11, and products is 10
C2H5OH + 3O2 --> 2CO2 + 3H2O
Given two masses from two reactants,
there is bound to be a limiting reactant
that runs out before the other... and
the unused, or excess reactant.
Grams to moles:
212g x 1 mol C2H5OH/ 46.07g = 4.60 mol
115g x 1 mol O2/32g = 3.59 mol
Since molar ratio (coefficients)
between C2H5OH and O2 is 1:3,
4.6 x 3 = 13.8 moles O2 needed
to completely combust
But there's only 3.59 moles O2..
so limiting reactant = O2
a) Mass H2O (water vapour produced):
Molar ratio O2 & H2O 3:3 = 1:1 same
3.59 mol x 18.0g/mol = 64.6g produced
b) O2
c) 64.6g/ 68g x 100 = 95% yield
e) 3.59/3 = 1.20 moles C2H5OH used
4.60 - 1.20 = 3.40 moles excess
3.40 mol x 46.07g/mol = 157g