A
What mass of KCl is needed to precipitate the silver ions from 26.0 mL of 0.100 M AgNO3 solution?
KCl + AgNO3 ==> AgCl + KNO3
mols AgNO3 = M x L = ?
moles KCl = moles AgNO3 from the 1:1 ratio in the equation.
g KCl = moles KCl x molar mass KCl.
32.2
To calculate the mass of KCl needed to precipitate the silver ions from the AgNO3 solution, we need to determine the mole ratio between KCl and AgNO3.
The balanced chemical equation for the reaction between KCl and AgNO3 is:
AgNO3 + KCl → AgCl + KNO3
From the balanced equation, we can deduce that one mole of AgNO3 reacts with one mole of KCl to form one mole of AgCl.
Step 1: Determine the moles of AgNO3
Moles = Molarity * Volume
= 0.100 M * 0.0260 L
= 0.00260 moles AgNO3
Step 2: Determine the moles of KCl
Since the mole ratio between AgNO3 and KCl is 1:1, the moles of KCl will also be 0.00260 moles.
Step 3: Determine the molar mass of KCl
The molar mass of KCl is calculated by adding the atomic masses of potassium (K) and chlorine (Cl):
Molar mass KCl = (1 mol K * atomic mass K) + (1 mol Cl * atomic mass Cl)
= (1 * 39.10 g/mol) + (1 * 35.45 g/mol)
= 74.55 g/mol
Step 4: Calculate the mass of KCl
Mass = moles of KCl * molar mass of KCl
= 0.00260 moles * 74.55 g/mol
= 0.193 g
Therefore, the mass of KCl needed to precipitate the silver ions from 26.0 mL of 0.100 M AgNO3 solution is 0.193 grams.
To determine the mass of KCl needed to precipitate the silver ions from the AgNO3 solution, we need to use the concept of stoichiometry and the balanced chemical equation.
The balanced chemical equation for the reaction between AgNO3 and KCl is:
AgNO3 + KCl -> AgCl + KNO3
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of KCl to produce 1 mole of AgCl.
Given:
Volume of AgNO3 solution (V) = 26.0 mL = 0.0260 L
Concentration of AgNO3 (C) = 0.100 M
To determine the moles of AgNO3 present in the solution, we can use the formula:
moles of AgNO3 = concentration (M) × volume (L)
moles of AgNO3 = 0.100 M × 0.0260 L
Now, we know that the stoichiometry between AgNO3 and KCl is 1:1. Therefore, the moles of KCl needed to precipitate all the silver ions would also be equal to moles of AgNO3.
Hence, moles of KCl = moles of AgNO3 = 0.100 M × 0.0260 L
To calculate the mass of KCl, we need to use the molar mass of KCl. The molar mass of KCl is the sum of the atomic masses of potassium (K) and chlorine (Cl). It is approximately 74.55 g/mol.
mass of KCl = moles of KCl × molar mass of KCl
mass of KCl = (0.100 M × 0.0260 L) × 74.55 g/mol
Finally, we can substitute the values to calculate the mass of KCl needed:
mass of KCl = (0.100 M × 0.0260 L) × 74.55 g/mol
Evaluating the above expression will give you the mass of KCl needed to precipitate the silver ions from the AgNO3 solution.