In the reaction C3H8 + 5O2---yields---3CO2 + 4H2O

If 20.0g of C3H3 and 20.0g of O2 are reacted, how can I calculate the # of moles of CO2 that can be produced?

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if 20.0g of C3H3 and 20.0g of O2 are reacted, how can I calculate the # of moles of CO2 that can be produced?

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* I think you meant 20.0 g of C3H8. Right?

a. Figure out the molecular weight (aka molecular mass) of C3H8 (propane).
b. Figure out how many moles of propane you started with (how many moles of C3H8 in 20.0g).
c. Figure out the molecular weight of O2
d. Figure out how many moles of O2 you started with (how many moles of O2 in 20.0g)

e. From the equation, above (you balanced it correctly), you need five times the number of moles of O2 as you do propane. So, from the answers to c and d, figure out which molecule is in excess and which molecule is limiting -- oxygen or propane.

f. All of the molecules of the limiting chemical (O2 or C3H8) will react but only the fraction of the molecules of the other chemical will react. The number of moles that will react is found from the chemical equation.

case 1. If oxygen is the limiting reagent, then divide the number of moles of oxygen by 5 to figure out the number of moles of C3H8 that will react. Do you see where the 5 comes from?
case 2. If propane is the limiting reagent, then multiply the number of moles of propane by 5 to figure out the number of moles of oxygen that will react. Do you see where the 5 comes from?

g. For each mole of propane that reacts, you will produce 3 moles of CO2. That's the final calculation.