Question

Calcium sulfate (gypsum) provides the orange color frequently assoicated with fireworks . The UNBALANCED equation is shown below. Balance the equation

____CaCl2+____Al2(SO4)3-->____CaSO4+____AlCl3

A) Write the complete ionic equation

B) Write the net ionic equation

C) What are the spectator ions in this reaciton.

D) A chemist decides to make her own calcium sulfate for a 4th of July party. If she has 527mL of 1.25M CaCl2 and 328 mL of 1.12M Al2(SO4) what is the theoretical yield for calcium sulfate in grams? (Molar mass calcium sulfate 136.14g/mole)

E) What ions will still be in solution after the reaction?

F) The reaction proceeded with 42% yield. How many grams of calcium sulfate does she have for her fireworks?

Answer 1

Balanced equation:
3 CaCl₂ + Al₂(SO₄)₃ ====> 3 CaSO₄ + 2 AlCl₃
Reaction type: double replacement

complete ionic equation

3 Ca²⁺ + 3Cl^- + 2Al³⁺+ 3SO₄^2- ====> 3 CaSO₄ +  2Al³⁺ + 3Cl^-

Calcium sulfate is insoluble. Al³⁺andSO₄^2-are spectator ion

Net ionic equation, cancel out ions which are identical on both sides which leaves:
3 Ca²⁺ + 3SO₄^2- ===> 3 CaSO₄

Question D

527mL of 1.25M CaCl2 = 527 x 1.25 / 1000 = 0.65875 Moles

328 mL of 1.12M Al2(SO4) = 328 x 1.12 /1000 = 0.36736 Moles

Limiting reagent = CaCl2

Moles of CaCl2 to be produced =   0.65875 Moles

Mass of CaCl2 =   0.65875 x 136.14 = 89.68 gm

Question E

Al₂(SO₄)₃ will be after reaction(50.56gm )

Question F

Amount of CaSO4 after reactio for 42% yield = 42 x 89.68 /100 =37.66 gm