2) (19 pts) Answer the following questions for the reaction of magnesium chlorid

Question

2) (19 pts) Answer the following questions for the reaction of magnesium chloride solution, MgCl2 (aq), and silver nitrate solution, AgNOs (aq) A) (3 pts) Write and balance a molecular equation that describes this reaction. Incli the physical states for all reactants and products. B) (3 pts) What is the complete/total ionic equation (TIE) for this reaction C) 3 pts) Give the net ionic equation (NIE) for this reaction D) (2 pts) Identify the spectator ion(s) for this reaction. E) (5 pts) Determine the theoretical yield, in grams, when 24.0 g of magnesium chloride is reacted with 18.0 g of silver nitrate? F) (I pt) What is the limiting reagent? G) (2 pts) Determine the actual yield if the percent yield of the insoluble product was 75.3 %.

Explanation / Answer

Magnesium Chloride         MgCl2

Silver nitrate                    AgNO3

a) MgCl2(aq) + 2 AgNO3(aq) ---------------------- Mg(NO3)2 (aq) + 2 AgCl(s)

b) Mg+2(aq) + 2 Cl-(aq) + 2 Ag+(aq) + 2 NO3-(aq) --------------- Mg+2(aq) + 2 NO3-(aq) + 2 AgCl(s)

c)The net ionic equation is 2 Ag+(aq) + 2 Cl-(aq) -------------- 2 AgCl(s)

d) The ions which are present on bothsides with out any change are called spectator ions.

In this equaion the spectator ions are Mg+2 ,and NO3-.

e)

mass of MgCl2 =24.0 grams

molar mass of MgCl2 = 95g

number of moles of MgCl= 24.0/95= 0.253 moles

mass of AgNO3 = 18.0 grams

molar mass of AgNO3= 169.87 gram/mole

number of moles of AgNo3 = 18.0/169.87 = 0.106 moles

according to equation

1 mole of MgCl2 = 2 mole of AgNO3

0.253 moles of MgCl2 = ?

                                      = 2x0.253/1 = 0.506 moles of AgNO3

we need 0.506 moles of AgNO3. but we have 0.106 moles of AgNO3.i. e AgNO3 is usedup first inthe reaction.

Hence the Limiting reagent is AgNO3.

according to equation

2 moles of AgNO3 = 2 moles of AgCl

0.106 moles of AgNO3 = 2x0.106/2 = 0.106 moles

number of moles of AgCl formed = 0.106 moles

molar mass of AgCl= 143.32 gram/mole

mass of 0.106 moles of AgCl= 0.106x143.32 = 15.19 grams

Theoritical yield of AgCl= 15.19 grams

F) AgNO3 is limiting reagent

G

Percent yiled = 75.3%

percent yield = Actual yiled/Theoritical yiled x100

Actual yiled = Percent Yield X Theoritical yield/100 = 75.3 x 15.19/100 = 11.43 grams

Actual yiled = 11.43 grams.

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