Mass-Mass Calculation Example Two: Potassium Iodide Reacts with Antimony(IV) Chloride



How many grams of antimony(III) chloride, SbCl(s), can be precipitated from solution when 15.7g of potassium iodide, KI(aq), in solution is consumed? 

The first step is to convert potassium iodide’s gram quantity to moles of potassium iodide, KI(aq), in solution. The second step uses the stoichiometry of the balanced equation

2KI(aq)+1SbCl(aq) 1SbI(s)+2KCl(aq)+ 1I(s)

 to find the moles of antimony(III) chloride, SbCl(s),  precipitated from solution. In the third step the molar amount of antimony(III) chloride, SbCl(s), produced is then converted to grams of antimony(III) chloride, SbCl(s), precipitated from solution.

 Step 1:  One mole of potassium iodide, KI, is 166.00 grams/mole and is found using the sum of the gram atomic mass of potassium(39.10g) and iodine(126.90g) from the periodic table. The formula mass calculation is as follows.

Element   Subscript   X   Element Mass = Formula Mass

    K                1           X      39.10g            =   39.10g/m

    I                 1           X      126.90g          =  126.90g/m

The molar mass of potassium iodide, KI, is 39.10g + 126.90g = 166.00g/m.

 Dividing the given 15.7 grams potassium iodide, KI(aq), in solution by the 166.00 grams/mole for the potassium iodide, KI, produces moles of potassium iodide, KI(aq), in solution consumed during the reaction.

15.7g KI ÷ (166.00g KI/1m KI)= 0.0946m KI

15.7 grams potassium iodide, KI, is equal to 0.0946 moles of potassium iodide, KI(aq), in solution.

Step 2: The 0.0946m moles of potassium iodide, KI(aq), in solution must be converted to moles antimony(III) chloride, SbCl (s), precipitated from solution using the balanced equation

 2KI(aq)+1SbCl (aq)1SbI(s)+2KCl(aq)+1I(s)

 0.0946m KI(aq) X (1mSbCl(s)/2mKI(aq)= 0.0473m SbCl(s)

Step 3: The 0.0473m antimony(III) chloride solid, SbCl(s), must be converted to grams. The formula mass for antimony(III) chloride, SbCl(s),  is found by the following formula mass calculation using gram atomic masses for each antimony(121.76g) and chlorine(35.45g) from the periodic table.

Element   Subscript    X   Element Mass = Formula Mass

    Sb               1           X       121.76g          =    121.76g

    Cl               3            X        35.45g           =    106.35g

The molar mass of antimony(III) chloride, SbCl(s), is 121.76g + 106.35g= 228.11g/m.

0.0473m SbCl (s) X (228.11g/1m SbCl(s)) = 10.8g  SbCl(s)

Therefore, 10.8g  antimony(III) chloride, SbCl(s), is produced at the same time 15.7g of potassium iodide in solution, KI(aq), is consumed    



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