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.

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.

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2