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 H2Oposted by Devron
28.36posted by Carmen