Question

Propionoic acid, C3H6O2, reacting with methanol will produce C4H8O2 and water (equation shown). 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

Please show work

Answer 1

When quantities for both reactants are listed, usually it is a limiting reagent (LR) problem. These are just two stoichiometry problems put together. I do these the long way but the long way is easier for me to explain. Remember this is just like a stoichiometry problem with an added twist.
C3H6O2 + CH4O → C4H8O2 + H2O
1. Determine moles of reactants.
a. mols C3H6O2 = g/molar mass = 70/74 = approx 0.95
b. mols CH4O = 60/32 = approx 1.9
2. Using the coefficients in the balanced equation, convert mols of 1a and 1b to mols of any product. I'll use C4H8O2.
2a. 0.95 mols C3H6O2 x (1 mol C4H8O2/1 mol C3H6O2) = about 0.95
2b. mols CH4O x (1 mol C4H8O2/1 mol CH4O) = about 1.9
In LR problems the lowest number wins. The reagent producing that number is the LR.

Answer 2

Provide the complete produce methanol Process Flow Diagram and explain the concept of stoichiometry including the calculations involved in the process (Precisely for the distillation portion).

Answer 3

Give the detailed steps involved in material balance To produce methanol. Assumptions of data should be provided