Question

The procedure called for reacting 2.00 g of sodium benzoate (13.9 mmol) with 5 mL of...

The procedure called for reacting 2.00 g of sodium benzoate (13.9 mmol) with 5 mL of 3M HCl.

(a) Calculate how many mmols of HCl was added if exactly 5 mL of the acid was used.

(b) Based on the amounts of sodium benzoate and HCl used, which reactant is the limiting reagent?

Homework Answers

Answer #1

a. number of moles of HCl = Molarity * volume

                                      = 3M * (5*10-3 lit) = 15 * 10-3 moles = 15 mmoles

b.

given number of moles of sodium benzoate = 13.9 mmol.

HCl is more than sodium benzoate.

So sodium benzoate is limiting reactant according to balanced chemical reaction.

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