Question

Calculate the percentage Yield of the preparation of Tris(thiourea)copper(I)chloridePreparation of Tris(Thiourea) Copper (I) Chloride Dissolve 5.0g of Thiourea in 25 mL of hot Water, add 1.0g of Copper Turnings. In a fume hood, add 5 mL of concentrated HCl to this mixture. Heat this on a Steam bath while the Copper dissolves. Do not boil. Add Water if needed to maintain the liquid at a constant level. Filter the hot solution and allow to cool slowly. White opaque crystals will form as the solution cools. Filter the crystals and wash them with Acetone. Allow the crystals to Air dry. Weigh the product and store in a labelled 3 dram vial. If the mass of the product obtained is 4.532g

Answer 1

In this case, it's neccesary to know which one is the limiting reactang between copper and thiourea. Let's determine the moles:
moles Cu = 1 / 63.55 = 0.0157 moles
moles (CH₄N₂S) = 5 / 76.12 = 0.0657 moles

The overall reaction would be:
Cu + 3CH₄N₂S-----------> Cu(SCN₂H₄)₃Cl

There's a 1:3 relation so:
moles of thiourea needed: 0.0157 * 3 / 1 = 0.0471 moles of Thiourea needed and we have 0.0657 so, the limiting reactant is the Cu.

The mass of product yielded would be:
m = 313.05 * 0.0157 = 4.9149 g

The %yield would be:
% = 4.532 / 4.9149 * 100
% = 92.2%

Hope this helps