What mass of KCl \rm KCl is needed to precipitate the silver ions from 13.0mL mL of 0.200 M M AgNO 3 \rm AgNO_3 solution
AgNO3(aq) + KCl(aq) ==> AgCl(s) + KNO3(aq)
mols AgNO3 = M x L = ?
mols KCl = the same (Look at the coefficients in the balanced equation.)
g KCl = mols KCl x molar mass KCl.
8.34
To determine the mass of KCl needed to precipitate the silver ions from the given AgNO3 solution, we need to use stoichiometry and the concept of limiting reactants.
The balanced chemical equation for the reaction between AgNO3 and KCl is:
AgNO3 + KCl → AgCl + KNO3
From the equation, we can determine that 1 mole of AgNO3 reacts with 1 mole of KCl to give 1 mol of AgCl precipitate.
Step 1: Calculate the number of moles of AgNO3 in the given volume and concentration.
Volume of AgNO3 solution = 13.0 mL
Concentration of AgNO3 solution = 0.200 M
To find the moles of AgNO3, we can use the following formula:
moles = concentration x volume (in liters)
moles of AgNO3 = 0.200 M * 0.0130 L = 0.0026 moles
Step 2: Determine the stoichiometric ratio between AgNO3 and KCl.
From the balanced equation, the ratio between AgNO3 and KCl is 1:1. This means that 1 mole of AgNO3 reacts with 1 mole of KCl.
Step 3: Determine the mass of KCl needed.
Since the stoichiometric ratio is 1:1, the number of moles of KCl needed is also 0.0026 moles.
To find the mass of KCl, we need to use its molar mass. The molar mass of KCl is 74.55 g/mol.
mass = moles x molar mass
mass of KCl = 0.0026 mol * 74.55 g/mol ≈ 0.192 g
Therefore, approximately 0.192 grams of KCl is needed to precipitate the silver ions from 13.0 mL of a 0.200 M AgNO3 solution.