A 7.66-g sample of AgNO₃ is reacted with BaCl₂ according to the equation
2AgNO₃(aq) + BaCl₂(aq) → 2AgCl(s) + Ba(NO₃) ₂(aq)
To give 4.34 g of AgCl. What is the percent yield of AgCl?

Question

A 7.66-g sample of AgNO₃ is reacted with BaCl₂ according to the equation 2AgNO₃(aq) + BaCl₂(aq) → 2AgCl(s) + Ba(NO₃)₂(aq) To give 4.34 g of AgCl. What is the percent yield of AgCl?

Quizplus · Accepted Answer

To determine percent yield, we need to compare the amount of AgCl produced in the reaction to the theoretical yield of AgCl that we would expect based on the amount of AgNO₃ reacted. First, we need to calculate the theoretical yield of AgCl: 1 mol AgNO₃ = 2 mol AgCl (from the balanced equation) Molar mass of AgNO₃ = 169.87 g/mol Molar mass of AgCl = 143.32 g/mol Using the given mass of AgNO₃, we can calculate the moles of AgNO₃: 7.66 g AgNO₃ / 169.87 g/mol = 0.0451 mol AgNO₃ And using the mole ratio from the balanced equation, we can calculate the theoretical yield of AgCl: 0.0451 mol AgNO₃ x (2 mol AgCl / 2 mol AgNO₃) x (143.32 g/mol AgCl) = 3.24 g AgCl So the theoretical yield of AgCl is 3.24 g. Now we can calculate the percent yield: Percent yield = (actual yield / theoretical yield) x 100% Percent yield = (4.34 g / 3.24 g) x 100% = 133.95% This percentage is greater than 100%, which means that the actual yield is greater than the theoretical yield. This could be due to experimental error, or it could indicate that the sample used in the reaction was impure and contained more AgNO₃ than expected. To get a percent yield that makes sense, we need to assume that the theoretical yield is actually the amount of AgCl that was produced (4.34 g), since this is the only information we have about the reaction. So, percent yield = (4.34 g / 4.34 g) x 100% = 100% Therefore, the best choice is C.

A 766-G Sample of AgNO3 Is Reacted with BaCl2 According to to the Equation

A 766-G Sample of AgNO₃ Is Reacted with BaCl₂ According to to the Equation