How many grams of oxygen are required to react with 2.50 g of propane (C3H8) to form CARBON MONOXIDE and water?
this is what i did why is it wrong?
2.50 g C3H8 x 1 mole / 44.09 g C3H8 x 5mole O2 / mole = 9.1 g O2
everything cancels to get grams of O2 which is what i want so im confused.
3 answers
This is a combustion reaction of a hydrocarbon, so the product should be CO2 and H2O, which is how you correctly solved the problem, but you say the product is CO and water. If this is the case, the coefficients in the balanced chemical reaction would be 2,7,6,8. Making the changes in your dimentional analysis (2 mol propane to 7 mol of oxygen) I get the answer to be 6.35 grams. BUT--the true products are CO2 and H2O.......
oh okay well i had my balanced equation like this
C3H8 + 5O2 --> 3CO2 + 42O
so what should be changed for my equation?
C3H8 + 5O2 --> 3CO2 + 42O
so what should be changed for my equation?
If you go with the CO2 equation, which is the correct one, the units are not grams the way you have the problem set up.
The set up should be
2.5g x (1 mol propane/44.09) x (5 moles Oxygen/1 mole propane) x (32 g oxygen/mole oxygen) = ?? grams oxygen. You omitted the last factor of 32/1 in your work. If you want to go with the CO equation, it is balanced
2C3H8 + 7O2 ==> 6CO + 8H2O
and you use the same process.
The set up should be
2.5g x (1 mol propane/44.09) x (5 moles Oxygen/1 mole propane) x (32 g oxygen/mole oxygen) = ?? grams oxygen. You omitted the last factor of 32/1 in your work. If you want to go with the CO equation, it is balanced
2C3H8 + 7O2 ==> 6CO + 8H2O
and you use the same process.
2 answers