100 ml of 15 g/L silver nitrate reacts with an unknown concentration solution of copper (ll) chloride. the reaction for me only 2 grams of silver (l) chloride. If 75 mL of the copper (ll) chloride solution was mixed in, what is the initial concentration of the solution?
One of the most poorly worded questions I've seen in some time.
From the description this is a limiting reagent problem in which the CuCl2 is the LR.
2AgNO3 + CuCl2 ==> 2AgCl + Cu(NO3)2
mols AgCl formed = grams/molar mass = ?
mols CuCl2 initially = 1/2 mols AgCl.
(CuCl2) = mols CuCl2/0.175L = ?
To find the initial concentration of the copper (II) chloride solution, we can set up a balanced equation for the reaction between silver nitrate and copper (II) chloride:
2AgNO₃ + CuCl₂ → 2AgCl + Cu(NO₃)₂
From the equation, we can see that 2 moles of silver nitrate react with 1 mole of copper (II) chloride to produce 2 moles of silver chloride.
Given that 100 mL of a 15 g/L silver nitrate solution reacts to form 2 grams of silver chloride, we need to convert the mass of silver chloride to moles. The molar mass of AgCl is 143.32 g/mol, so 2 grams of AgCl is equal to:
2 g / 143.32 g/mol = 0.0139 moles of AgCl
Since 2 moles of AgNO₃ react to produce 2 moles of AgCl, we can determine the number of moles of AgNO₃ that reacted:
0.0139 moles AgCl / 2 moles AgNO₃ = 0.00695 moles AgNO₃
Now, we can determine the number of moles of CuCl₂ based on the stoichiometry of the balanced equation. Since the ratio of AgNO₃ to CuCl₂ is 2:1, the number of moles of CuCl₂ reacted is also 0.00695 moles.
The volume of the CuCl₂ solution used is given as 75 mL. To find the initial concentration of the CuCl₂ solution, we need to convert the volume to liters:
75 mL = 75/1000 L = 0.075 L
Finally, we can calculate the initial concentration of the CuCl₂ solution (in g/L) by dividing the number of moles of CuCl₂ by the volume in liters:
Initial concentration = (0.00695 moles) / (0.075 L) = 0.093 g/L
Therefore, the initial concentration of the copper (II) chloride solution is 0.093 g/L.