Question

Propionoic acid, C3H6O2, reacting with methanol, CH4O, will produce C4H8O2 and water. If 70.0 g of propionoic acid and 60.0 g of methanol are reacted together, which one will be the limiting reactant?

C3H6O2 + CH4O → C4H8O2 + H2O

Answer 1

First, find the molar masses of each substance.

C3H6O2 = 74 g/mol

CH4O = 32 g/mol

Then, use the molar mass to find how many moles of each are present:

C3H6O2:
`(70.0g\ C_(3)H_(6)O_(2))((1 mol\ C_(3)H_(6)O_(2))/(74.0 g\ C_(3)H_(6)O_(2))) = 0.945945945 mol\ C_(3)H_(6)O_(2)`

CH4O:
`(60.0g\ CH_(4)O)((1mol\ CH_(4)O)/(32g\ CH_(4)O)) = 1.875mol\ CH_(4)O`

There is a higher number of moles of CH4O than C3H6O2, so the C3H6O2 will run out faster. Therefore, C3H6O2 is the limiting reactant.