Answer to Question #154045 in Chemistry for Tailey Zhang

Question

if u are given 877 ml of silver nitrate at a concentration of (0.64mol)/(l), how much precipitate will form?

Answer 1

The number of the moles of silver nitrate in 877 mL of 0.64 M solution is:

$n = cV = 0.64·877·10^{-3} = 0.561$ mol.

Usually, silver nitrate forms precipitates with the halogen salts, for example:

AgNO₃ + NaCl $\rightarrow$ AgCl$\downarrow$ + NaNO₃.

In these reactions, 1 mole of precipitate is formed when 1 mole of silver nitrate is formed. Therefore, 0.561 mol of the precipitate is formed.

Finally, if the chloride is used as an anion, the mass of the precipitate (when an excess of NaCl is used) will be:

$m = n·M = 0.561·143.32 = 80.44$ g.

In this equation , the molar mass of AgCl is used, 143.32 g/mol.