In a government lab, 827 g of chlorobenzene (C6H5Cl) is reacted with 345 g of chloral (C2HOC13). If the actual yield of DDT (C14H9C15) is 464 g, then what is the percent yield?
2C6H5CI+CHOC13 --> C14H2Cl5 + H20

Percent yield = 55.9%

Explanation:

To solve this question we must find the moles of the reactants in order to find limting reactant. Using the chemical equation and limiting reactant we can find the moles of DDT assuming a 100% of yield (Theoretical Yield). As percent yield is:

Percent Yield = Actual yield (464g) / Theoretical Yield * 100

We can find percent yield as follows:

Moles chlorobenzene- Molar mass: 112.56g/mol-

827g * (1mol / 112.56g) = 7.35 moles

Moles chloral -Molar mass: 147.3877g/mol-

345g * (1mol / 147.3877g) = 2.34 moles

For a complete reaction of 2.34 moles of chloral are required:

2.34moles CHOCl3 * (2mol C6H5Cl / 1mol CHOCl3) = 4.68 moles Chlorbenzene. As there are 7.35 moles, Chlorobenzene is the excess reactant and chloral the limiting reactant.

The theoretical moles of DDT produced = Moles DDT = 2.34 moles. The mass is (Molar mass DDT: 354,49 g/mol):

2.34mol * (354.49g / mol) = 830g DDT = Theoretical mass

Percent yield = Actual yield (464g) / Theoretical Yield (830g) * 100

it's all four options.