How much AgCl is isolated using this and Ag(NO3)?

Question

Chad decides to sneak into Ms. Smith's class room. He finds bottles of solution containing silver. He realizes that if he extracts the silver, he will strike it rich! First, he sees silver nitrate. He also realizes that AgCl is an insoluble compound

a) He finds a bottle labeled ".75M MgCl2", but only 300mL are remaining. How much AgCl is isolated using this and Ag(NO3)?

Balanced Equation:

Dimensional Analysis:

Theoretical Yield:

b) Draw the particles after the reaction is complete and the particles have settled.

c) % yield is 55%. How much AgCl is isolated?

Answer 1

The unbalanced equation is:

 

AgNO₃ + MgCl₂ ---> AgCl + Mg(NO₃)₂

 

Balancing it is pretty simple...

 

2AgNO₃ + MgCl₂ ---> 2AgCl + Mg(NO₃)₂

 

If there are 300 ml of .75M MgCl₂, that means that there are .3 * .75 = .225 moles of MgCl₂, since 300 ml = .3 l.

 

Since there are 2 molecules of AgCl produced for every molecule of MgCl₂, that means that .45 moles of AgCl will be produced.

 

So the theoretical yield in grams will be .45 times the molecular mass of AgCl.

 

The molecular mass of AgCl is:

 

Ag -- 107.87

Cl -- 35.45

TOTAL -- 143.32

 

So the theoretical yield is .45 * 143.32 = 64.49 grams of AgCl.

 

If the actual yield is 55%, then .55 * 64.49 = 35.47 grams of AgCl will be produced.

 

Sorry, but the diagram can't be done here.