Assume the molar mass of the triacylglycerol (vegetable oil) in the fatty acid s

Question

Assume the molar mass of the triacylglycerol (vegetable oil) in the fatty acid salt (soap) synthesis given above is 885 g/mol. If the reaction is run with 3.01 g of triacylglycerol, 4.00 mL water and 0.3501 g of NaOH, what is the limiting reagent (the reactant that limits the amount of product formed because it is in the shortest supply)?

OAAWWFAAWW 3 NaOH —O Molar Mass: 885.45 g/mol Molar Mass: 40.00 g/mol Triacylglycerol (Vegetable oil) Sodium Hydro Na — -OH Molar Mass: 306.46 g/mol -OH Na OH Molar Mass: 304.44 g/mol Molar Mass: 92.09 g/mol Nat Glycerol Molar Mass: 302.43 g/mol Fatty Acid Salts (Soap)

Explanation / Answer

first calculate the no of moles of triacylglycerol = weight / molar mass = 3.01 g / 885.45 g/mol

= 0.003399 moles

no of moles of NaOH = 0.3501 g / 40 g/mol = 0.00875 mol

from the given equation

one mole of triacylglycerol required 3 moles of NaOH

accordingly

0.003399 moles of triacylglycerol required 3 x 0.003399 = 0.01012 moles of NaOH

but actually we have 0.00875 moles of NaOH

so limiting agent is NaOH

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