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A 45.00-g sample of silver nitrate is mixed with 55.00 g of hydrochloric acid to form a white precipitate of silver chloride. After the solution is filtered and dried, a white precipitate of mass 33.50 g is collected.
a. Determine the limiting reactant.
b. Determine the theoretical yield of silver chloride.
c. Determine the percent yield of silver chloride.



Answer :

Oseni

a. The limiting reactant is silver nitrate

b. The theoretical yield of silver chloride is 37.94 grams

c. The percent yield of silver chloride is 88.3%

Stoichiometric problem

First, let us see the equation of the reaction:

[tex]HCl + AgNO_3 -- > HNO_3 + AgCl[/tex]

The mole ratio of the reactants is 1:1.

Mole of 45 g sample of silver nitrate = 45/170

                                                = 0.2647 moles

Mole of 55 g sample if hydrochloric acid = 55/36.46

                                                = 1.5068 moles

Thus, the limiting reactant is silver nitrate.

Mole ratio of silver nitrate and silver chloride = 1:1.

Equivalent mole of silver chloride = 0.2647 moles

Mass of 0.2647 moles silver chloride = 0.2647 x 143.32

                                                              = 37.94 grams

Theoretical yield of silver chloride = 37.94 grams

The percentage yield of silver chloride = 33.5/37.94 x 100

                                                         = 88.3%

More on stoichiometric problems can be found here: https://brainly.com/question/14465605

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