A 0.5566-g sample containing CuCl2 . 2H2O (MW = 170.47) was dissolved in 25.00 mL of...

A 0.5566-g sample containing CuCl₂ . 2H₂O (MW = 170.47) was dissolved in 25.00 mL of water. About 25.00 mL of a 10% aqueaous solution of thioacetamide was added to the solution and heated. The black CuS precipitate formed upon heating the solution was filtered, dried, and weighed. If the precipitate was found to weigh 0.2074 g, calulate the percentage of CuCl₂ . 2H₂O in the sample.

Solutions

Expert Solution

given sample wt = 0.5566g

after treatment with thioacetamide black CuS is ppted, &found 0.2074g

% yield = amount produced exptlly / amount for 100% yield * 100

= 0.2074 /0.5566 * 100

= 37.26%

thank u

queen_honey_blossom answered 2 years ago

Next > < Previous

Related Solutions

A 125.0 mg sample of methylamine (CH3NH2 MW = 31.06) was dissolved in 25.00 mL of...

A 125.0 mg sample of methylamine (CH3NH2 MW = 31.06) was dissolved in 25.00 mL of carbonate-free water. {The Ka for CH3NH3+ is 2.3 x 10-11.} Determine the pH of the resulting solution after the addition of each of the following solutions. A) 25.00 mL of pure carbonate-free water B) 25.00 mL of 0.1609 M aqueous methylamine hydrochloride solution C) 25.00 mL of 0.0512 M aqueous HCl solution

4.3120 g KNO3, 1.2345 g of CuCl22H2O, and 0.5678 g of BaCl22H2O are used to prepare...

4.3120 g KNO3, 1.2345 g of CuCl2*2H2O, and 0.5678 g of BaCl2*2H2O are used to prepare K2CuBa(NO2)6. Calculate the theoretical yield of K2CuBa(NO2)6 in grams.

A 5.59 g sample of a solid containing Ni is dissolved in 20.0 mL water. A...

A 5.59 g sample of a solid containing Ni is dissolved in 20.0 mL water. A 5.00 mL aliquot of this solution is diluted to 100.0 mL and analyzed in the lab. The solution was determined to contain 5.70 ppm Ni. Determine the molarity of Ni in the 20.0 mL solution. Determine the mass in grams of Ni in the original sample. Determine the weight percent of Ni in the sample.

A sample containing a mixture of SrCl2·6H2O (MW = 266.62 g/mol) and CsCl (MW = 168.36...

A sample containing a mixture of SrCl2·6H2O (MW = 266.62 g/mol) and CsCl (MW = 168.36 g/mol) originally weighs 1.7215 g. Upon heating the sample to 320 °C, the waters of hydration are driven off SrCl2·6H2O, leaving the anhydrous SrCl2. After cooling the sample in a desiccator, it has a mass of 1.2521 g. Calculate the weight percent of Sr, Cs, and Cl in the original sample.

A 100.0 mL solution containing 0.830 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.830 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.298 M KOH. Calculate the pH of the solution after the addition of 48.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. pH = At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the fully...

A 100.0 mL solution containing 0.8366 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.8366 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.3203 M KOH. Calculate the pH of the solution after the addition of 45.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the fully protonated, intermediate,...

A 100.0 mL solution containing 0.7590 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.7590 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.2906 M KOH. Calculate the pH of the solution after the addition of 45.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. pH=? At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM-, and M^2-, which represent the fully protonated,...

A 100.0 mL solution containing 0.753 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.753 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.259 M KOH. Calculate the pH of the solution after the addition of 50.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the fully protonated, intermediate,...

A 100.0 mL solution containing 0.756 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.756 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.217 M KOH. Calculate the pH of the solution after the addition of 60.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. At this pH (pH = 9.48), calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the...

A 100.0 mL solution containing 0.976 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.976 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.296 M KOH. Calculate the pH of the solution after the addition of 56.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27 . pH = At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the...

Subjects