A8.53 g sample of bacl2 barium chloride is reacted with silver nitrate to produce silver chloride and bariums nitrate to give 8.72 g of silver chloride. write the balanced reaction, and find the theoretical yield and the percent yield of silver chloride.

BaCl2 + 2AgNO3  ⇒ 2AgCl + Ba(NO3)2

Theoretical yield: 11.74g AgCl

Percent yield: 74.27%

Explanation:

BaCl2 + 2AgNO3  ⇒ 2AgCl + Ba(NO3)2

The percent yield = actual yield/ theoretical yield * 100%

Actual yield: The amount of product actually obtained from a reaction, in this case is 8.72g AgCl

Theoretical yield: The amount of product that would result if all the limiting reagent reacted. Now, we have the amount of BaCl2, so BaCl2 is our limiting reagent, because the maximum amount of product formed depends on how much of this reactant was present. Taking into consideration the balanced reaction, we can calculate the theoretical yield, so.

mass molar BaCl2 = 208.23 g/mol

mass molar AgCl = 143.32 g/mol

First,  

moles of BaCl2= 8.53 g BaCl2*( 1mol BaCl2/ 208.23g BaCl2) = 0.04096mol

so, we have 0.04096 mol BaCl2, given that one mol of BaCl2 reacts to produce 2 mol of AgCl (watch the balanced reaction above) , so, the moles of AgCl are twice the moles of BaCl2,  we should get 0.0819 mol AgCl and 11.74 g AgCl

g AgCl= 0.0819mol AgCl* (143.32g AgCl/ 1mol AgCl) = 11.74g AgCl theoretical yield

% yield = 8.72 g AgCl/ 11.74g AgCl* 100= 74.27%

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