A triglyceride of molar mass $890 g mo l^{- 1}$ is completely saponified with NaOH according to $triglyceride + 3 NaOH glycerol + 3 soap$ . Taking the molar mass of NaOH as $40 g mo l^{- 1}$ , calculate the mass of NaOH needed for $75.0 g$ of the triglyceride. Give the answer in grams to two decimal places.

Numerical answer type, answer in g, tolerance 0.01.

Answer: 10.11 g

One mole of triglyceride consumes 3 mol NaOH. Moles of triglyceride = 75.0/890 = 0.08427 mol, so NaOH = 3 x 0.08427 = 0.25281 mol. Mass of NaOH = 0.25281 x 40 = 10.11 g.

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A triglyceride of molar mass 890 g mol−1 is completely saponified with NaOH according to triglyceride+3NaOH​glycerol+3soap. Taking the molar mass of NaOH as 40 g mol−1, calculate the mass of NaOH needed for 75.0 g of the triglyceride. Give the answer in grams to two decimal places.

Answer: 10.11 g

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