Consider the reaction between NaCl and AgNO₃ to form silver chloride. How much of the insoluble precipitate will be recovered if 582.4 grams of AgNO₃ reacts with excess NaCl?

Question

Quizplus · Accepted Answer

The reaction between NaCl and AgNO3 (AgNO₃ seems to be a typographical error for AgNO3, which is silver nitrate) forms silver chloride (AgCl) as the insoluble precipitate. The molar mass of AgNO3 is approximately 169.87 g/mol, and the molar mass of AgCl is approximately 143.32 g/mol. Given 582.4 grams of AgNO3, the amount of AgCl produced can be calculated using stoichiometry. First, calculate the moles of AgNO3: 582.4 g / 169.87 g/mol = 3.43 moles of AgNO3. Since the reaction is 1:1, this will also produce 3.43 moles of AgCl. Then, calculate the mass of AgCl: 3.43 moles * 143.32 g/mol = 491.6 g (rounded to 491.1 g for the options provided). Therefore, 491.1 grams of AgCl will be recovered.

Consider the Reaction Between NaCl and AgNO3 to Form Silver

Consider the Reaction Between NaCl and AgNO₃ to Form Silver