C3H8 + O2 CO2 + H2O
If you start with 14.8g of C3H8 and 3.44g of O2, how many grams of H2O are produced?
Which is the limiting reagent (which reactant created less H2O? _______
How many grams of the non-limiting reagent are actually used?
How many grams of the non-limiting reagent are in excess?
Note : Balance first the equation
It becomes C3H8 +5O2 3 CO2 + 4H2O
Molar mass of
Molar mass of O
Molar mass of H
Moles of C3H8
Moles of O
Grams of H2O produced
Limiting reagent is O2
Grams of C3H8 used
Excess of C3H8
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