What mass KCl is needed to precipitate AgCl from 15.0 mL of .200 M AgNO3
AgNO3 + KCl ==> AgCl + KNO3
mols AgNO3 = M x L = ?
mols KCl needed = mols AgNO3 because of the coefficients in the balanced equation.
mols KCl = grams KCl/molar mass KCl. You know molar mass and mols KCl, solve for grams.
To determine the mass of KCl needed to precipitate AgCl from 15.0 mL of 0.200 M AgNO3, we need to use stoichiometry and the balanced equation for the reaction between AgNO3 and KCl.
The balanced equation for the reaction is:
AgNO3 + KCl → AgCl + KNO3
From the equation, we can see that one mole of AgNO3 reacts with one mole of KCl to produce one mole of AgCl. Therefore, the molar ratio between AgNO3 and KCl is 1:1.
First, let's calculate the number of moles of AgNO3 that we have:
Moles of AgNO3 = concentration of AgNO3 × volume of AgNO3 solution
Moles of AgNO3 = 0.200 M × 0.0150 L
Moles of AgNO3 = 0.00300 mol
Since the molar ratio between AgNO3 and KCl is 1:1, we have 0.00300 moles of KCl needed to react with 0.00300 moles of AgNO3.
Now, we need to calculate the mass of KCl using its molar mass. The molar mass of KCl is 74.55 g/mol.
Mass of KCl = moles of KCl × molar mass of KCl
Mass of KCl = 0.00300 mol × 74.55 g/mol
Mass of KCl = 0.22365 g
Therefore, approximately 0.224 g of KCl is needed to precipitate AgCl from 15.0 mL of 0.200 M AgNO3.