A 50.00-mL sample of 0.0250 M silver nitrate is mixed
with 0.0400 M chromium(III) chloride.
(a) What is the minimum volume of chromium(III)
chloride required to completely precipitate silver chloride?
(b) How many grams of silver chloride are produced
from (a)?

1. The basic reaction is Ag+ + Cl- => AgCl
2. Wie have 50 ml of a 0,025 Mol/l silver nitrate solution .
This correspond  to 0,05 l*0,025 mol/l = 0,00125 mol = 1,25 mmol Ag+
The same amount we need  for chloride
3. The Chromium-III-chloride solution is 0,04 M
CrCl3 =>  Cr3+ + 3 Cl-. The molarity  regarding chloride is 3 times = 0,12 M = 0,12 mol/ l = 0,12 mmol/ ml
To get 1,25 mmol Cl- we need 1,25mmol/ 0,12 mmol/ ml = 10,41 ml
We produce 1,25 mmol AgCl what correspond to the mass of 179,15 mg with molar mass of 143,32 g/mol

Quote from: Harini on October 03, 2022, 05:16:36 AM
A 50.00-mL sample of 0.0250 M silver nitrate is mixed
with 0.0400 M chromium(III) chloride.
(a) What is the minimum volume of chromium(III)
chloride required to completely precipitate silver chloride?
(b) How many grams of silver chloride are produced
from (a)?

Write the balanced equation first.
3Ag NO₃ +CrCl₃====>3AgCl(s) + Cr(NO₃)₃
It is clear from the equation that 3 moles of silver nitrate are reacting with one mole of chromium (lll) chloride .
Moles of AgNO₃ = Mx V (L)
= 0.050 x  0,025 moles
Now moles of CrCl₃ needed to precipitate Ag+ =( 0.050 x  0,025 )?3. moles. = Mx V (L)
Calculate volume  of CrCl₃ from above equation.
B) From balanced equation  we know
3 moles of AgNO₃ is giving 3 moles  of AgCl
Convert moles of AgNO₃ to AgCl and convert it to grams by multiplying it with molar mass of AgCL