Assume the molar mass of the triacylglycerol (vegetable oil) in the biofuel synt

Question

Assume the molar mass of the triacylglycerol (vegetable oil) in the biofuel synthesis given above is 885 g/mol. Sodium hydroxide (NaOH) is not a reactant, rather it serves as a catalyst for the reaction. Methanol (CH3OH) has a density of 0.791 g/mL. If the reaction is run with 5.00 g of triacylglycerol (vegetable oil), 1.00 mL methanol and 0.020 g of NaOH (40.00 g/mol), what is the limiting reagent (the reactant that limits the amount of product formed because it is in the shortest supply)?

A) Triacylglycerol (vegetable oil)

B) Fatty acid methyl esters (biofuel)

C) Glycerol

D) Sodium hydroxide (NaOH)

E) Methanol

F) Cannot be determined from the information provided

Explanation / Answer

moles of triacylglycerol=mass/molar mass=5.00 g/885g/mol=0.0056 moles

moles of methanol in 1 ml =0.791g/32.04g/mol=0.025 moles

moles of NaOH=0.020g/40.00g/mol=0.005 moles

The reactants react in the ratio- triacylglycerol : methanol=1:3(3 moles of catalyst is needed)

so moles of methanol required to react with 0.0056 moles of triacylglycerol=3*0.0056=0.017 moles.

The amount of triacylglycero is less than that is needed to react with the given amount of methanol,So triacylglycerol is the imiting reagent.

Answer A

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