how many grams of O2 will it take to Oxidize 88 grams of C3H8 to CO and H2O
To determine the amount of O2 required to oxidize 88 grams of C3H8 (propane) to CO (carbon monoxide) and H2O (water), we first need to balance the equation for the combustion reaction:
C3H8 + O2 → CO + H2O
Balancing the equation, we get:
C3H8 + 5O2 → 3CO + 4H2O
From the balanced equation, we can see that 1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO and 4 moles of H2O.
Next, we need to convert the mass of C3H8 to moles. The molar mass of C3H8 is:
(3 * atomic mass of carbon) + (8 * atomic mass of hydrogen) = (3 * 12.01 g/mol) + (8 * 1.008 g/mol) = 44.1 g/mol
Therefore, 88 grams of C3H8 is equal to:
88 g * (1 mol / 44.1 g) = 2 moles of C3H8
Using the stoichiometry of the balanced equation, we can determine the moles of O2 required to react with 2 moles of C3H8:
2 moles of C3H8 * (5 moles of O2 / 1 mole of C3H8) = 10 moles of O2
Finally, we need to convert moles of O2 to grams. The molar mass of O2 is 32 g/mol, so:
10 moles of O2 * (32 g / 1 mol) = 320 grams of O2
Therefore, it will take 320 grams of O2 to oxidize 88 grams of C3H8 to CO and H2O.
Follow the steps in this example.
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