Answer: 0.0185 grams of NaCI.
Step-by-step explanation:
To determine the amount of NaCl required to precipitate most of the Ag+ ions from the given AgNO3 solution, we can use the following steps:
Determine the number of moles of Ag+ ions in the solution:
Molarity of AgNO3 solution = 1.17 x 10^-2 M
Volume of AgNO3 solution = 2.7 x 10^-2 L
Number of moles of Ag+ ions = Molarity x Volume
Number of moles of Ag+ ions = (1.17 x 10^-2) x (2.7 x 10^-2)
Number of moles of Ag+ ions = 3.159 x 10^-4 moles
Use the stoichiometry of the balanced chemical equation to determine the amount of NaCl required to precipitate most of the Ag+ ions:
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
From the balanced chemical equation, we can see that 1 mole of AgNO3 reacts with 1 mole of NaCl to produce 1 mole of AgCl.
Therefore, the number of moles of NaCl required to precipitate most of the Ag+ ions is also 3.159 x 10^-4 moles.
Convert the moles of NaCl to grams:
Molar mass of NaCl = 58.45 g/mol
Mass of NaCl = Number of moles of NaCl x Molar mass of NaCl
Mass of NaCl = (3.159 x 10^-4) x 58.45
Mass of NaCl = 0.0185 grams
Therefore, 0.0185 grams of NaCl are required to precipitate most of the Ag+ ions from 2.7 x 10^-2 L of the 1.17 x 10^-2 M AgNO3 solution.
Bear in mind that this is not a rocket science. But it was fun. I enjoy the process. Thank you for asking.