Consider the following reaction:
How many grams of water will be produced if the reaction is conducted with 75.6g of C4H10O and 95.5 g of oxygen?
Solution This solution is FREE courtesy of BrainMass!
First you need to calculate the moles of reactants.
Molecular weight of C4H10O = 4*12+10+16 =74 (g/mol)
Therefore the mole of C4H10O is = 75.6/74 = 1.02 (mol)
Molecular weight of O2 = 32 (g/mol)
Therefore the mole of O2 is =95.5/32 = 2.98 (mol)
According to the reaction
C4H10O + 6O2-> 4CO2+5H2O
This reaction tells you that 1 mole of C4H10O needs to react completely with 6 moles of O2.
Here we have 1.02 mole of C4H10O, this means we will need more than 6 moles of O2 to consume all the C4H10O. However we only have 2.98 moles of O2, therefore O2 is the limiting reactant. I hope the concept of limiting reactant is not too harsh for you.
So therefore the mole of C4H10O consumed will be only 1/6 of the moles of O2
Now we can calculate the mole of H2O produced =5*0.5=2.5 (mol H2O)
The mass of H2O produced is 2.5*18 = 45 (g).