What mass of silver chloride can be produced from 1.21 L of a 0.200 M solution of silver nitrate?
To determine the mass of silver chloride that can be produced, we first need to determine the amount of silver nitrate present in the solution.
Using the formula:
M1V1 = M2V2
where M1 is the molarity of the initial solution, V1 is the volume of the initial solution, M2 is the molarity of the final solution, and V2 is the volume of the final solution.
Plugging in the values:
(0.200 M)(1.21 L) = M2(unknown volume)
M2 = 0.200 x 1.21 / 1 = 0.242
This means that the final molarity of silver nitrate is 0.242 M. Since silver nitrate is a 1:1 stoichiometry with silver chloride, the molarity of silver chloride is also 0.242 M.
Now we can calculate the moles of silver chloride produced:
1.21 L x 0.242 mol/L = 0.29302 mol
Finally, we can convert moles to grams using the molar mass of silver chloride (143.32 g/mol):
0.29302 mol x 143.32 g/mol = 42.0 g
Therefore, 42.0 grams of silver chloride can be produced from 1.21 L of a 0.200 M solution of silver nitrate.