To assay silver content in an alloy, a 13.38 g sample of alloy was dissolved in nitric acid. The addition of 0.53 molar sodium chloride resulted in the precipitation of silver chloride. If the alloy were pure silver, what volume of sodium chloride is needed to precipitate the silver?
Convert 13.38 g Of Ag to moles by dividing by the gram-atomic mass of Ag.
1 mole of Ag produces 1 mole of AgNO3
1 mole of AgNO3 + 1 mole NaCl ---> 1 mole AgCl
so.....
Moles of Ag = Moles of AgNO3 = moles of NaCl
(Moles of NaCl) / (molarity of NaCl) = liters NaCl
Just a quick note here to have you realize that this is a ridiculous problem. An alloy is at least two somethings together so it can't be pure silver as the problem states.
To find the volume of sodium chloride needed to precipitate the silver, we need to use stoichiometry and the balanced chemical equation that describes the reaction between silver and sodium chloride.
The balanced chemical equation for the reaction is:
Ag + NaCl → AgCl + Na
From the equation, we can see that one mole of silver reacts with one mole of sodium chloride to produce one mole of silver chloride.
First, convert the mass of the alloy sample to moles of silver. To do this, we need the molar mass of silver. The molar mass of silver is approximately 107.87 g/mol.
Number of moles of silver = mass of alloy sample / molar mass of silver
Number of moles of silver = 13.38 g / 107.87 g/mol
Now, we can use stoichiometry to determine the moles of sodium chloride needed to react with the silver. From the balanced equation, we know that 1 mole of silver reacts with 1 mole of sodium chloride.
Number of moles of sodium chloride = Number of moles of silver
Number of moles of sodium chloride = 13.38 g / 107.87 g/mol
Finally, we can find the volume of sodium chloride using the molarity of the sodium chloride solution.
Number of moles of sodium chloride = Molarity of sodium chloride solution * Volume of sodium chloride (in liters)
Volume of sodium chloride (in liters) = Number of moles of sodium chloride / Molarity of sodium chloride solution
In this case, the molarity of sodium chloride is given as 0.53 M.
Volume of sodium chloride (in liters) = (13.38 g / 107.87 g/mol) / 0.53 mol/L
After calculating the expression, you will get the volume of sodium chloride needed to precipitate the silver in liters.