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
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?
What is the experimental yield (in g of precipitate) when 15.9 mL of a 0.6 M...
What is the experimental yield (in g of precipitate) when 15.9 mL
of a 0.6 M solution of iron(iii) chloride is combined with 19.9 mL
of a 0.738 M solution of lead(ok) nitrate at a 76.6% yield?
What is the experimental yield (in g of precipitate) when 15.9 mL of a 0.6 M...
What is the experimental yield (in g of precipitate) when 15.9 mL
of a 0.6 M solution of iron(iii) chloride is combined with 19.9 mL
of a 0.738 M solution of lead(ok) nitrate at a 76.6% yield?
What is the experimental yield (in g of precipitate) when 17.6 mL of a 0.7 M...
What is the experimental yield (in g of precipitate) when 17.6
mL of a 0.7 M solution of iron(III) chloride is combined with 17.4
mL of a 0.613 M solution of silver nitrate at a 81.2% yield?
What is the theoretical yield (in g of precipitate) when 18.4 mL
of a 0.73 M solution of sodium phosphate is combined with 18 mL of
a 0.695 M solution of aluminum chloride?
What is the experimental yield (in g of precipitate) when 18.3 mL of a 0.679 M...
What is the experimental yield (in g of precipitate) when 18.3 mL
of a 0.679 M solution of barium hydroxide is combined with 18.8 mL
of a 0.648 M solution of aluminum nutrate at a 88.4% yield?
What is the experimental yield (in g of precipitate) when 19.8 mL of a 0.663 M...
What is the experimental yield (in g of precipitate) when 19.8
mL of a 0.663 M solution of barium hydroxide is combined with 19.2
mL of a 0.727 M solution of aluminum nitrate at a 77.4% yield?
What is the experimental yield (in g of precipitate) when 16.6 mL of a 0.5 M...
What is the experimental yield (in g of precipitate) when 16.6
mL of a 0.5 M solution of sodium chloride is combined with 17.2 mL
of a 0.567 M solution of silver nitrate at a 89.8% yield?
What is the theoretical yield (in g of precipitate) when 17.6 mL
of a 0.501 M solution of iron(III) chloride is combined with 17.9
mL of a 0.614 M solution of lead(II) nitrate?
Nitrogen and oxygen react to form nitric oxide (NO). All...
40. What is the experimental yield (in g of precipitate) when 17.2 mL of a 0.503...
40. What is the experimental yield (in g of precipitate) when
17.2 mL of a 0.503 M solution of barium hydroxide is combined with
15.4 mL of a 0.556 M solution of aluminum nitrate at a 76.9%
yield?
What is the theoretical yield (in g of precipitate) when 17.6 mL of a 0.501 M...
What is the theoretical yield (in g of precipitate) when 17.6 mL
of a 0.501 M solution of iron(III) chloride is combined with 17.9
mL of a 0.614 M solution of lead(II) nitrate?
Nitrogen and oxygen react to form nitric oxide (NO). All
substances are in the gas phase. If 0.448 atm of nitrogen and 0.371
atm of oxygen react, what is the partial pressure of nitric oxide
(in mmHg) when this reaction goes 74.5 complete. The temperature
and volume...
What is the theoretical yield (in g of precipitate) when 19.8 mL of a 0.694 M...
What is the theoretical yield (in g of precipitate) when 19.8 mL
of a 0.694 M solution of iron(III) chloride is combined with 16.1
mL of a 0.568 M solution of lead(II) nitrate?
ADVERTISEMENT
ADVERTISEMENT
Latest Questions
- An investor is considering purchasing a Treasury bond with a 20 year maturity, an 8% coupon...
- Write a program that copies the contents of one file to a destination file. This program...
- how to conduct a marketing mix for 3d crest white strips
- Is big data analytics used for service innovation at Amazon? If so, please write a brief...
- A shrink-fit process has been used to enhance the pressure carrying capacity of a cylinder. During...
- Ten-year zero coupon bonds issued by the U.S. Treasury have a face value of $1,000 and...
- The Blossom Department of Transportation has issued 25-year bonds that make semiannual coupon payments at a...
ADVERTISEMENT