What mass of NaCl is required to precipitate all the Ag ions from 20 mL of .100 M AgNO3 solution
How many mols Ag^+ do you have?
mols Ag^+ = M x L = ?
mols NaCl = mols Ag^+ (look at the coefficients in NaCl + AgNO3 =>AgCl + NaNO3.
mass NaCl = mols NaCl x molar mass NaCl
To find the mass of NaCl required to precipitate all the Ag ions from the AgNO3 solution, we need to determine the stoichiometry of the reaction between NaCl and AgNO3, specifically the molar ratio between NaCl and AgNO3.
From the balanced equation of the reaction:
2AgNO3 + NaCl -> 2AgCl + NaNO3
we can see that 2 moles of AgNO3 react with 1 mole of NaCl to form 2 moles of AgCl.
Now, let's calculate the moles of AgNO3 present in the solution:
moles of AgNO3 = volume of solution (in L) x concentration of AgNO3 (in M)
= 0.020 L × 0.100 mol/L
= 0.002 mol
Since the molar ratio between AgNO3 and NaCl is 2:1, we can say that 0.002 moles of AgNO3 will react with 0.001 moles of NaCl (half the amount).
Finally, to determine the mass of NaCl required, we multiply the number of moles of NaCl by the molar mass of NaCl:
mass of NaCl = moles of NaCl × molar mass of NaCl
= 0.001 mol × 58.44 g/mol (molar mass of NaCl)
= 0.05844 g
Therefore, approximately 0.05844 grams of NaCl is required to precipitate all the Ag ions from a 20 mL, 0.100 M AgNO3 solution.