(1)When aqueous solutions of sodium iodide, NaI, and lead(II) acetate, Pb(C2H3O2
ID: 885470 • Letter: #
Question
(1)When aqueous solutions of sodium iodide, NaI, and lead(II) acetate, Pb(C2H3O2)2, are mixed, which of the following correctly describes the outcome?
(a)No reaction occurs when these two solutions are mixed.
(b)A precipitate of sodium acetate, NaC2H3O2, is formed when the solutions are mixed.
(c)A precipitate of lead(II) iodide, PbI2, is formed when the solutions are mixed.
(d)These two aqueous solutions cannot form a precipitate because both reactants are soluble.
(2)A mixture of 54.0 g of S and 1.06×102 g of Cl2 reacts completely to form S2Cl2 and SCl2.
Find the mass of S2Cl2 formed.
m = g
Explanation / Answer
1)
(c) A precipitate of lead(II) iodide, PbI2, is formed when the solutions are mixed.
NaI(aq) + Pb(CH3O2)2(aq) -------------> PbI2 (s) + CH3COONa(aq)
2)
3S + 2Cl2 --------------> S2Cl2 + SCl2
No.of moles of S = 54 / 32 = 1.68 mol
No.of moles of Cl2 = 106 / 71 = 1.5 mol
So, Sulfur is limiting reactants.
2 moles of Sulfur will form 1 mole of S2Cl2
1.68 moles of Sulfur will form 0.56 moles of S2Cl2.
Therefore, mass of S2Cl2 = 0.56 x (64+71) = 75.6 g
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