Answer to Question #136614 in General Chemistry for Miça

Question

PROBLEM:
An aqueous solution of 0.00585 mol FeCl3 is mixed with an aqueous solution of 0.700 g
NaOH to form a precipitate according to:
FeCl3(aq) + NaOH(aq)  Fe(OH)3(s) + Na +(aq) + Cl-(aq)
a. How many grams of precipitate form if the yield of the reaction is 75.2%?
b. Assuming theoretical yield (100% yield), how many moles of Na+(aq) that
remains in solution?
a. How many grams of precipitate form if the yield of the reaction is 75.2%?

Answer 1

FeCl3(aq) +3 NaOH(aq) --> Fe(OH)3(s) + 3Na +(aq) +3 Cl-(aq)

Use this balanced equation to calculate yield

mols of FeCl3 = 0.00585 mol 

mols of NaOH present =0.700g/40.0 g/mol 

= 0.0175 mols

Since mols of NaOH is fewer than required mol ratio

==> limiting reactant is NaOH

a)

mols of Fe(OH)3 formed = 1 mol fe(OH)3/3 mol NaOH*0.0175 mols NaOH

 = 0.00583 mol

mass of Fe(OH)3 formed= 0.00583 mol * 106.87 g/mol 

 = 0.623 g

% yield = 75.2 %

thus actual yield = 75.2 /100 * 0.623 g =0.468 g

mass of precipitate = 0.468 g

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b)

number of mols of NaOH = number of mols of Na+ions

hence number of mols of Na+ = 0.0175 mols

ANSWER

a)mass of precipitate = 0.468 g

b)number of mols of Na+ = 0.0175 mols 

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