Ethanol, C2H6O, is added to gasoline to produce E85, a fuel for automobile engines. How many grams of 02 are required for the complete combustion of 421 g of ethanol?
And then it gives this equation: C2H5OH + 3O2 --> 2C02 +3H20
I don't even know where to begin.
And then it gives this equation: C2H5OH + 3O2 --> 2C02 +3H20
I don't even know where to begin.
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They are being sneaky...Ethanol, C2H6O this is the molecular formula, C2H5OH is the structural formula... the same thing.
C2H5OH + 3O2 --> 2C02 +3H20 is your recipe
421 g of ethanol* 3 O2 (32g)/C2H5OH (46g) = 879g O2
3 O2 (32g)/C2H5OH (46g) is the mole ratio and the molar mass combined
C2H5OH + 3O2 --> 2C02 +3H20 is your recipe
421 g of ethanol* 3 O2 (32g)/C2H5OH (46g) = 879g O2
3 O2 (32g)/C2H5OH (46g) is the mole ratio and the molar mass combined