Question

6. If 5.7 g of sodium acetate combined with 5.3 q of calcium hydroxide, write the chemical equation by predicting the product. a. Calculate the limiting reagent. (2 pt) b. Calculate the amount of sodium containing product is produced. (2 pt) c. Calculate the amount of excess. (2 pt) d. Calculate the percent yield of sodium containing product. (2 pt)

Answer 1

The reaction that occurs is:

2 NaC2H3O2 + Ca (OH) 2 = Ca (C2H3O2) 2 + 2 NaOH

The moles of the reagents are calculated:

n Acetate = g / MM = 5.7 / 82 = 0.07 mol

n Hydroxide = 5.3 / 74.1 = 0.07 mol

a) The limit reagent is acetate, since it requires two moles to react with one mole of hydroxide.

b) The mass of NaOH is calculated:

m NaOH = 0.07 mol acetate * (2 mol NaOH / 2 mol acetate) * (40 g NaOH / 1 mol) = 2.8 g

c) 0.035 mol of excess calcium hydroxide remain, the mass is calculated:

m hydroxide = 0.035 mol * 74.1 g / mol = 2.59 g

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