A 5.95-g sample of AgNO₃ is reacted with excess BaCl₂ according to the equation
2AgNO₃(aq) + BaCl₂(aq) $\to$ 2AgCl(s) + Ba(NO₃) ₂(aq)
To give 3.75 g of AgCl.What is the percent yield of AgCl?

Question

A 5.95-g sample of AgNO₃ is reacted with excess BaCl₂ according to the equation 2AgNO₃(aq)+ BaCl₂(aq)$\to$ 2AgCl(s)+ Ba(NO₃)₂(aq) To give 3.75 g of AgCl.What is the percent yield of AgCl?

Quizplus · Accepted Answer

The theoretical yield of AgCl can be calculated using stoichiometry. The molar mass of AgNO3 is approximately 169.87 g/mol, so 5.95 g of AgNO3 is 0.035 mol. According to the balanced equation, 2 moles of AgNO3 produce 2 moles of AgCl, so 0.035 mol of AgNO3 should produce 0.035 mol of AgCl. The molar mass of AgCl is approximately 143.32 g/mol, so the theoretical yield is 0.035 mol × 143.32 g/mol = 5.0162 g. The percent yield is (3.75 g / 5.0162 g) × 100% = 74.7%.

A 595-G Sample of AgNO3 Is Reacted with Excess BaCl2 According

A 595-G Sample of AgNO₃ Is Reacted with Excess BaCl₂ According