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
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.
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).
Give the detailed steps involved in material balance To produce methanol. Assumptions of data should be provided
To determine the limiting reactant, we need to compare the number of moles of each reactant to see which one will be completely consumed.
First, let's calculate the number of moles of each reactant:
Molecular weight of propionic acid (C3H6O2) = (3 * atomic weight of carbon) + (6 * atomic weight of hydrogen) + (2 * atomic weight of oxygen)
= (3 * 12.01 g/mol) + (6 * 1.008 g/mol) + (2 * 16.00 g/mol)
= 72.06 g/mol
Number of moles of propionic acid (C3H6O2) = Mass / Molecular weight
= 70.0 g / 72.06 g/mol
= 0.970 mol
Molecular weight of methanol (CH4O) = (1 * atomic weight of carbon) + (4 * atomic weight of hydrogen) + (1 * atomic weight of oxygen)
= (1 * 12.01 g/mol) + (4 * 1.008 g/mol) + (1 * 16.00 g/mol)
= 32.04 g/mol
Number of moles of methanol (CH4O) = Mass / Molecular weight
= 60.0 g / 32.04 g/mol
= 1.874 mol
Now, we can compare the number of moles of each reactant. The propionic acid has 0.970 moles, and the methanol has 1.874 moles.
To determine the limiting reactant, we compare the coefficient ratios of the balanced equation. The coefficient ratio between the propionic acid and methanol is 1:1.
Since the coefficient ratio is 1:1, we can conclude that the limiting reactant is propionic acid because it has fewer moles (0.970 mol) compared to methanol (1.874 mol). The reaction will be limited by the amount of propionic acid.