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 3.57 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)→ 2AgCl(s)+ Ba(NO₃)₂(aq) To give 3.57 g of AgCl.What is the percent yield of AgCl?

Quizplus · Accepted Answer

To calculate the theoretical yield of AgCl, we need to first calculate the moles of AgNO₃ used. Molar mass of AgNO₃ = 169.87 g/mol 5.95 g of AgNO₃ is equal to 0.035 moles. From the balanced equation, we know that two moles of AgNO₃ produce two moles of AgCl. Therefore, the theoretical yield of AgCl is 0.035 x 2 x 143.32 g/mol = 10.00 g. Percent yield = (actual yield / theoretical yield) x 100% Substituting the given values, we get: Percent yield = (3.57 g / 10.00 g) x 100% = 35.7% Rounding off the answer to one decimal place, the correct choice is C) 71.1%.

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

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