From the question I assume we know that O2 is the limiting reagent and butane is the excess reagent.
mols butane = grams/molar mass = ?
mols O2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols butane. That will give you the mols butane used by the O2. Subtract initial mols butane - mols butane used by the O2 reaction = mols butane left. Then convert mols butane to grams by g butane = mols butane x molar mass butane.
Gaseous butane
CH3CH22CH3
will react with gaseous oxygen
O2
to produce gaseous carbon dioxide
CO2
and gaseous water
H2O
. Suppose 36.0 g of butane is mixed with 41. g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction.
1 answer
1 answer