posted by Cliff .
Please help me solve this problem.
I don't know how to approach this problem type.
C3H8(g) + 5 O2 (g) ---- 3 CO2(g) + 4H2O(g)
Conside the reaction. If you start with 7.0 moles of C3H8 (propane)and 7.0 moles O2, ________ mole(s) of carbon dioxide will be produced.
Thanks for your help.
When BOTH reactants are given, we know it is a limiting reagent type problem.
How many moles CO2 will be produced with 7.0 moles C3H3 and an excess of oxygen? That will be 7.0 moles C3H8 x (3 moles CO2/1 mole C3H8) = 21.0 moles CO2.
How many moles CO2 will be produced with 7.0 moles O2 and an excess of C3H8. That will be 7.0 moles O2 x (3 moles CO2/5 moles O2) = 4.2 moles CO2.
The answers don't agree which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller one and the reactant producing that value is the limiting reagent.
Remember how to do this type problem. I expect you will see a number of them.