Question

There are many minerals which contain silver. A 2.00 g sample of one of these, stephanite, is placed in water and all of its silver ions are released. It takes 2.89 mL of 2.19 M magnesium chloride to precipitate all the silver ions just released as the solid silver chloride. Calculate the mass percent of silver in stephanite to the correct number of significant figures. Assume that stephanite contains no chloride ions.

Answer 1

Answer:

68.3%

Explanation:

First, let us look at the equation of reaction involving silver and magnesium chloride:

2Ag + MgCl2 ----> 2AgCl + Mg

1 mole of MgCl2 is required to precipitate 2 moles of Ag completely from the solution. That is a ratio of 1 to 2.

Now, mole of MgCl2 used to precipitate all the Ag

             = molarity x volume

                      = 2.19 M x 2.89/1000

                         = 0.0063291 mole

Since 1 mole of MgCl2 would always require 2 moles of Ag, 0.0063291 mole will therefore require:

      0.0063291 x 2 = 0.0126 mole of Ag

This means that 0.0126 mole of Ag is present in stephanie.

Mass of silver in stephanie = mole x molar mass

                              = 0.0126 x 107.8682

                                      = 1.365 g

Thus, 1.365 g of silver is present in 2.00 g sample of stephanie.

Mass percent of silver in stephanie = 1.365/2.00 x 100

                                         = 68.25% = 68.3% to the correct number of significant figure.