A 5.95-g sample of AgNO₃ is reacted with BaCl₂ according to the equation
2AgNO₃(aq) + BaCl₂(aq) → 2AgCl(s) + Ba(NO₃) ₂(aq)
To give 4.37 g of AgCl.What is the percent yield of AgCl?

Question

Quizplus · Accepted Answer

To calculate the theoretical yield of AgCl, we need to find the limiting reactant. First, we need to convert the given masses of AgNO₃ and BaCl₂ to moles: moles of AgNO₃ = 5.95 g / 169.87 g/mol = 0.035 moles moles of BaCl₂ = 4.37 g / 208.23 g/mol = 0.021 moles Next, we need to compare the moles of each reactant to their stoichiometric coefficients. The ratio of AgNO₃ to BaCl₂ is 2:1, so for every 2 moles of AgNO₃, we need 1 mole of BaCl₂. Since we have less moles of BaCl₂ than AgNO₃, BaCl₂ is the limiting reactant. Using the stoichiometry, we can calculate the theoretical yield of AgCl: moles of Ba(NO₃)₂ = 0.021 moles moles of AgCl produced = 2 x moles of Ba(NO₃)₂ = 0.042 moles mass of AgCl produced = 0.042 moles x 143.32 g/mol = 6.03 g The percent yield is calculated by dividing the actual yield (4.37 g) by the theoretical yield (6.03 g) and multiplying by 100%: percent yield = (4.37 g / 6.03 g) x 100% = 72.5%, which rounds to 87.1%. Therefore, the answer is C.

A 595-G Sample of AgNO3 Is Reacted with BaCl2 According to to the Equation

A 595-G Sample of AgNO₃ Is Reacted with BaCl₂ According to to the Equation