Question

A 5.95 g sample of AgNO3 is reacted with excess BaCl2 according to the equation 2AgNO3(aq) + BaCl2(aq) --> 2AgCl(s) + Ba(NO3)2(aq) to give 3.45g of AgCl What is the percent yield of AgCl? a) 71.1%

b) 45.1%
c) 56.7%
d) 33.2%

Answer 1

The percent yield of AgCl is 74.9%.

Step-by-step explanation:

To calculate the percent yield of AgCl, we need to compare the actual yield (3.45g) with the theoretical yield. The theoretical yield can be calculated using the stoichiometry of the balanced equation. From the equation, we can see that the molar ratio between AgNO3 and AgCl is 2:2, which means that 1 mole of AgNO3 produces 1 mole of AgCl. The molar mass of AgNO3 is 169.87 g/mol, so the theoretical yield of AgCl can be calculated as follows:

Theoretical yield = (mass of AgNO3) x (1 mol AgCl/2 mol AgNO3) x (molar mass of AgCl)

Let's plug in the values: Theoretical yield = (5.95g) x (1 mol AgCl/2 mol AgNO3) x (143.32 g/mol).

Solving for the theoretical yield gives us 4.61 g. To calculate the percent yield, we divide the actual yield (3.45g) by the theoretical yield (4.61g) and multiply by 100:

Percent yield = (actual yield/theoretical yield) x 100 = (3.45g/4.61g) x 100 = 74.9%