posted by Jefferson
How many grams of water will be produced in the complete combustion of 20.0g of Octane, C8H18(g) and 50.0g Oxygen gas?
Combustion reactions always follow this general reaction:
CxHy + O2 -----> CO2 + H2O
Plug in Octane into the the reaction and then balance:
C8H18 + O2 -----> CO2 + H2O
2 C8H18 + 25 O2 -----> 16 CO2 + 18 H2O
Just eyeballing it, octane is the limiting reagent.
****To determine the limiting reagent, you have to find the number of moles that each reactant can produce (i.e., which reactant will you run out of first.)
20.0g *(1 mole/114.23 g)= moles of C8H18
moles of C8H18*(18 mol of H2O/2 moles of C8H18)= moles of H2O
moles of H2O*(18.00g/mole)= mass of H2O