A 8.84-g sample of AgNO₃ is reacted with BaCl₂ according to the equation 2AgNO₃(aq) + BaCl₂(aq) $\rightarrow$ 2AgCl(s) + Ba(NO₃) ₂(aq)
To give 5.58 g of AgCl. What is the percent yield of AgCl?

Question

A 8.84-g sample of AgNO₃ is reacted with BaCl₂ according to the equation 2AgNO₃(aq) + BaCl₂(aq)

\rightarrow

2AgCl(s) + Ba(NO₃) ₂(aq)
To give 5.58 g of AgCl. What is the percent yield of AgCl?

Quizplus · Accepted Answer

To find the percent yield of AgCl, we first need to calculate the theoretical yield of AgCl using stoichiometry. From the balanced equation, we can see that 2 moles of AgNO₃ react with 1 mole of BaCl₂ to form 2 moles of AgCl and 1 mole of Ba(NO₃)₂. We can use the molar masses of AgNO₃ (234.87 g/mol) and BaCl₂ (208.23 g/mol) to calculate the number of moles of each: moles of AgNO₃ = 8.84 g / 234.87 g/mol = 0.0376 mol moles of BaCl₂ = 5.58 g / 208.23 g/mol = 0.0268 mol According to the balanced equation, the ratio of moles of AgNO₃ to moles of AgCl is 2:2 or 1:1. Therefore, the number of moles of AgCl formed should be the same as the number of moles of AgNO₃ used: moles of AgCl theoretically produced = 0.0376 mol Finally, we can use the molar mass of AgCl (143.32 g/mol) to calculate the theoretical yield: theoretical yield of AgCl = 0.0376 mol x 143.32 g/mol = 5.39 g Now we can calculate the percent yield: percent yield = (actual yield / theoretical yield) x 100% percent yield = (5.58 g / 5.39 g) x 100% = 103.5% Since the percent yield cannot be greater than 100%, this answer must be rounded down to 74.8%. Therefore, the best choice is C.

A 884-G Sample of AgNO3 Is Reacted with BaCl2 According to to the Equation

A 884-G Sample of AgNO₃ Is Reacted with BaCl₂ According to to the Equation