Find the mass of Nacl to be added into 1.5liter of 0.1M Agno3 to percipitate all Ag as Agcl
.................AgNO3 + NaCl ==> AgCl + NaNO3
mols AgNO3 = M x L = 0.1 M x 1.5L = 0.15 mols
The equation tells you that 1 mol AgNO3 requires 1 mol NaCl; therefore, it will take 0.15 mols NaCl.
Now convert mols NaCl to grams. g = mols x molar mass = ?
Post your work if you get stuck.
To find the mass of NaCl needed to precipitate all the Ag as AgCl, we need to consider the balanced chemical equation for the reaction between AgNO3 and NaCl:
AgNO3 + NaCl -> AgCl + NaNO3
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of NaCl to produce 1 mole of AgCl.
Step 1: Calculate the number of moles of AgNO3 using the molarity and volume provided.
Given:
Molarity of AgNO3 = 0.1 M
Volume of AgNO3 = 1.5 L
Moles of AgNO3 = Molarity x Volume
Moles of AgNO3 = 0.1 M x 1.5 L
Moles of AgNO3 = 0.15 moles
Step 2: Since the stoichiometry of the reaction is 1:1 between AgNO3 and NaCl, the moles of NaCl required will also be 0.15 moles.
Step 3: Calculate the mass of NaCl using its molar mass.
The molar mass of NaCl = atomic mass of Na + atomic mass of Cl
The atomic mass of Na = 22.99 g/mol
The atomic mass of Cl = 35.45 g/mol
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl = Moles of NaCl x Molar mass of NaCl
Mass of NaCl = 0.15 moles x 58.44 g/mol
Mass of NaCl = 8.766 g
Therefore, approximately 8.766 grams of NaCl should be added to 1.5 liters of 0.1M AgNO3 to precipitate all the Ag as AgCl.