Q, What volume of 0.472 mol/L AgNO3 will precipitate the chloride ion in 40ml of 0.183 mol/L AlCl3?
- Answer is 46.6ml, but how to solve this question?
Moles Ag= Volume*Molarity
But moles of Ag has to equal moles of Cl ion (it forms AgCl).
Moles Cl=.04*.184*3 (three ions in AlCl3)
set them equal, and solve for Volume
To solve this question, you need to use the concepts of stoichiometry and balanced chemical equations.
Step 1: Write a balanced chemical equation for the reaction between silver nitrate (AgNO3) and aluminum chloride (AlCl3).
2AgNO3 + 3AlCl3 → 3AgCl + 3Al(NO3)3
Step 2: Determine the stoichiometry of the reaction. From the balanced equation, you can see that it takes 2 moles of AgNO3 to react with 3 moles of AlCl3, resulting in the formation of 3 moles of AgCl.
Step 3: Calculate the moles of chloride ions (Cl-) in the 40 mL of 0.183 mol/L AlCl3 solution.
Moles of Cl- = concentration (mol/L) × volume (L)
Moles of Cl- = 0.183 mol/L × 0.040 L = 0.00732 mol Cl-
Step 4: Determine the volume of AgNO3 solution required to precipitate the chloride ions.
Using the stoichiometry, we know that 2 moles of AgNO3 react with 3 moles of AlCl3, resulting in the formation of 3 moles of AgCl. This means that for every 3 moles of AgCl formed, we need 2 moles of AgNO3.
Moles of AgNO3 required = (2 moles AgNO3 / 3 moles AgCl) × 0.00732 mol Cl-
Moles of AgNO3 required = 0.00488 mol AgNO3
Step 5: Calculate the volume of 0.472 mol/L AgNO3 solution required using the concentration.
Volume of AgNO3 = Moles of AgNO3 required / concentration (mol/L)
Volume of AgNO3 = 0.00488 mol AgNO3 / 0.472 mol/L = 0.01034 L = 10.34 mL
Therefore, the volume of 0.472 mol/L AgNO3 solution required to precipitate the chloride ions in 40 mL of 0.183 mol/L AlCl3 solution is approximately 10.34 mL. However, it's worth noting that the actual answer given is 46.6 mL, so there might be an additional factor or calculation involved in the given answer.