How do you find the percent yield from the below information? te ethanol. Dry th

Question

How do you find the percent yield from the below information? te ethanol. Dry the product using ethanol and finally with small portions and wash the bottle. Calculate the theoretical a uv dry, store it in a stoppered yield and determine your percentage yield. B. synthesis of Potassium Trioxalatochromate (llil Trihydrate (Kalcr(czolal. 3Hzo) The equation for the reaction performed in this synthesis is K2Cr2007 7H2C204 2H20 2K2C204.Hzo 6CO 17H2O into Add 11.1 g of oxalic acid dihydrate in 20.0 mL of cold water in a 100-mL beaker and stir this suspension 3.60 g of potassium dichromate. The orange-colored mixture should spontaneously warm up a t to boiling as a vigorous evolution of carbon dioxide commences. When the reaction has subsided, dissolve 4.20 g of potassium oxalate monohydrate in the hot green-black liquid. Cool the viscous reaction mixture in ice. Mix about 1 mL of this solution with 5 mL of 95% ethanol and use the pasty solid for seeding as described below. Add about 10 mL of 95% ethanol with stirring to the cooled reaction mixture in the beaker and introduce the pasty solid. The cooled mixture should soon thicken with crystals which should be filtered by suction with a Buchner funnel and filter flask after they have cooled in ice for 15-30 minutes. wash the crystals on the funnel with three 10-ml portions of 50% aqueous alcohol followed by 25 mL of 95% ethanol. Dry the product using a uv lamp. Weigh the dried material and store in a stoppered bottle. Calculate the theoretical yield and your percent yield.

Explanation / Answer

Use the balanced chemical equation as given:

K2Cr2O7 + 7 H2C2O4.2H2O + 2 K2C2O4.H2O <=====> 2 K3[Cr(C2O4)3].3H2O + 6 CO2 + 17 H2O

Find out the molar masses:

K2Cr2O7 = (2*39 + 2*52 + 7*16) g/mol = 294 g/mol.

H2C2O4.2H2O = (2*1 + 2*12 + 4*16 + 2*2*1 + 2*16) g/mol = 126 g/mol.

K2C2O4.H2O = (2*39 + 2*12 + 4*16 + 2*1 + 1*16) g/mol = 184 g/mol.

K3[Cr(C2O4)3].3H2O = (3*39 + 1*52 + 3*2*12 + 3*4*16 + 3*2*1 + 3*16) g/mol = 487 g/mol.

As per the balanced stoichiometric equation,

1 mole K2Cr2O7 = 7 moles H2C2O4.2H2O = 2 moles K2C2O4.H2O = 2 moles K3[Cr(C2O4)3].3H2O

Find out the moles of the reactants taken and the limiting reactant.

Mass of K2Cr2O7 = 3.6028 g; moles taken = (3.6028 g)/(294 g/mol) = 0.01225 mole.

Mass of H2C2O4.2H2O = 11.1031 g; moles taken = (11.1031 g)/(126 g/mol) = 0.08812 mole.

Mass of K2C2O4.H2O = 4.2066 g; moles taken = (4.2066 g)/(184 g/mol) = 0.02286 mole.

Now find out the limiting reactant. Find the ratio of the moles of the reactants.

K2Cr2O7:H2C2O4.2H2O:K2C2O4.H2O = 0.01225:0.08812:0.02286 = 1:7.19:1.866

Therefore, K2C2O4.H2O is the limiting reactant and the limiting reactant will form the product.

Moles of product formed = (0.02286 mole K2C2O4.H2O)*(2 moles product/2 moles K2C2O4.H2O) = 0.02286 mole.

Mass of K3[Cr(C2O4)3].3H2O formed = (0.02286 mole)*(487 g/mol) = 11.13282 g

Actual mass of K3[Cr(C2O4)3]3H2O obtained = (61.8194 – 58.0400) g = 3.7794 g.

Percent yield of product = (3.7994 g)/(11.13282 g)*100 = 33.9483 33.95% (ans).

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