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.
oh okay well i had my balanced equation like this
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.
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.......
Your calculations are incorrect because you used the wrong mole ratio between propane (C3H8) and oxygen (O2). To correctly determine the amount of oxygen required to react with 2.50 g of propane, you need to use the balanced chemical equation for the combustion reaction.
The balanced equation for the combustion of propane is:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the balanced equation, we can see that one molecule of propane (C3H8) reacts with five molecules of oxygen (O2) to produce three molecules of carbon dioxide (CO2) and four molecules of water (H2O).
To solve this problem, follow these steps:
Step 1: Calculate the number of moles of propane (C3H8) using its molar mass:
2.50 g C3H8 x (1 mole C3H8 / 44.09 g C3H8) = 0.0567 moles C3H8
Step 2: Use the mole ratio to find the number of moles of oxygen (O2) required:
0.0567 moles C3H8 x (5 moles O2 / 1 mole C3H8) = 0.2835 moles O2
Step 3: Calculate the mass of oxygen (O2) using its molar mass:
0.2835 moles O2 x (32.00 g O2 / 1 mole O2) = 9.072 g O2
Therefore, the correct answer is 9.072 grams of oxygen (O2) are required to react with 2.50 grams of propane (C3H8) to form carbon monoxide and water.