Question
In: Chemistry
What is the experimental yield (in g of precipitate) when 19.6 mL of a 0.6 M...
What is the experimental yield (in g of precipitate) when 19.6 mL of a 0.6 M solution of iron(III) chloride is combined with 16.4 mL of a 0.662 M solution of silver nitrate at a 75.4% yield?
Solutions
Expert Solution
balanced equation: FeCl3(aq) + 3AgNO3(aq) ----> Fe(NO3)3(aq) + 3AgCl(s)
no of mol of FeCl3 formed = M*V
= 19.6*0.6
= 11.76 mmol
no of mol of AgNO3 formed = 16.4*0.662
= 10.86 mmol
1 mol FeCl3(aq) = 3 mol AgNO3(aq)
no of mol of precipitate(AgCl) = 10.86 mmol
theoretical yield of AgCl formed(T.Y) = n*Mwt = 10.86*10^-3*143.32
= 1.56 g
percent yield of AgCl formed = (P.Y/T.Y)*100
75.4 = (x/1.56)*100
practical yield of yield of AgCl formed(P.Y) = 1.18 g
queen_honey_blossom answered 2 years agoRelated Solutions
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