What Mass of NaCl is required to precipitate all of the silver ions from 75.0 mL of a 0.10M solution of AgNO3?

Question

I need to do a Chemistry lab and cannot figure out how to even go about this question. Help Please!

Answer 1

First, you must know that 75 mL (0.075L) of a .1M AgNO3 solution contains .0075 mols of silver ions. AgNO3 reacts with NaCl in a 1:1 ratio to form the AgCl precipitate. And since the there are ~58.45 g NaCl / mol, you would need at least .439 grams of NaCl.

Note - If you are about to do a lab, and have no idea how to even get started with solving an elementary problem such as this, you should probably speak to the instructor for further clarification on stoichiometry before the lab. A chemistry lab can be a dangerous place if you don't know what you're doing.

Answer 2

Atomic weights: Na = 23 Cl = 35.5 NaCl = 58.5

AgNO3 + NaCl ===> AgCl + NaNO3

75.0mLsoln x 0.10molAgNO3/1000mLsoln x 1molNaCl/1molAgNO3 x 58.5gNaCl/1molNaCl = (75.0)(0.10)(58.5)/(1000) = 0.439 g NaCl

The first factor comes from the given amount of solution. The second factor comes from the molarity, 0.10M. The mL solution cancel, leaving moles of AgNO3. The third factor comes from the balanced equation. It may seem silly to make a big deal multiplying by 1/1, but it's important thinking through the problem to realize that the moles of AgNO3 cancel, leaving moles of NaCl. The fourth factor comes from the formula weight of NaCl. The moles of NaCl cancel, leaving grams NaCl.