Ammonium sulfate + Aluminum chloride Aluminum sulfate + ammonium chloride a) wri

Question

Ammonium sulfate + Aluminum chloride Aluminum sulfate + ammonium chloride

a) write and balance this equation.

B)If you started with 0.435 g of ammonium sulfate and reacted it with 1.13 g of aluminum chloride, how many grams of aluminum sulfate could be produced?

c) What was the limiting reactant in part b?

d) What was the excess reactant in part b?

e) How many grams of the excess reactant actually reacted in the reaction described in part b?

I really need to see the steps so I can understand how to do this please. thanks so much!!

Explanation / Answer

a. 3(NH4)2SO4(aq) + 2AlCl3(aq) ----------------> 6NH4Cl(aq) + Al2(SO4)3(aq)

b. no of moles of (NH4)2SO4   = W/G.M.Wt

                                              = 0.435/132   = 0.00329moles

   no of moles of AlCl3             = W/G.M.W

                                              = 1.13/133.34   = 0.00847moles

c.2 moles of AlCl3 react with 3 moles of (NH4)2So4

   0.00847 moles of AlCl3 react with = 3*0.00847/2   = 0.0127 moles of (NH4)2SO4 is required

(NH4)2SO4 is limiting reactant

3 moles of (NH4)2SO4 react with AlCl3 to gives 1 moles of Al2(SO4)3

0.00329 moles of (NH4)2SO4 react with AlCl3 to gives = 1*0.00329/3   = 0.0011 moles of Al2(SO4)3

mass of Al2(SO4)3   = no of moles * gram molar mass

                              = 0.0011*342   = 0.3762g

d. 3 moles of (NH4)2SO4 react with 2 moles of AlCl3

0.00329 moles of (NH4)2SO4 react with = 2*0.00329/3 = 0.00219 moles of AlCl3

AlCl3 is excess reactant

e. the no of moles of excess reactant AlCl3    = 0.00847-0.00219   = 0.00628 moles of excess

The amount of excess reactant AlCl3 = no of moles * gram molar mass

                                                         = 0.00628*133.34   = 0.837g of AlCl3

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