Answer:
5.48 g
Explanation:
M: 367.01
3Pb(NO3)2(aq) + 2CrBr3(aq) → 3PbBr2(s) + 2Cr(NO3)3(aq)
1. Moles of Pb(NO3)2
n = 30.0 mL × (0.551 mmol/1 mL) = 16.53 mmol Pb(NO3)2
2. Moles of PbBr2
n = 16.53 mmol Pb(NO3)2 × (3 mmol PbBr/3 mol Pb(NO3)2)
= 16.53 mmol PbBr2
(3) Mass of PbBr2 formed
m = 16.53 mmol × (367.01 mg/1 mol) = 6067 mg
(4) Mass of PbBr2 in solution
PbBr2 is slightly soluble, so a significant amount will remain in solution.
Its solubility is 973 mg/100 mL.
Mass of dissolved PbBr2 = 60.0 mL × 973 mg/100 mL = 584 mg
(5) Mass of precipitate
Mass of precipitate = mass of PbBr2 formed - mass of PbBr2 in solution
m = 6067 mg - 584 mg = 5480 mg = 5.48 g
The mass of precipitate is 5.48 g.
The mass of PbBr2 precipitate produced is 12.1 g.
The equation of the reaction is;
3Pb(NO3)2(aq) + 2CrBr3(aq) –>3PbBr2(s) + 2Cr(NO3)3(aq)
Number of moles of Pb(NO3)2 = 60.0/1000 L × 0.551 M = 0.0331 moles
From the information in the question, we already know that Pb(NO3)2 is the limiting reactant and CrBr3 is the reactant in excess.
From the equation of the reaction;
3 moles of Pb(NO3) yields 3 moles of PbBr2
0.0331 moles of Pb(NO3) yields 0.0331 moles × 3 moles/3 moles
= 0.0331 moles of PbBr2
Mass of PbBr2 = number of moles × molar mass
Mass of PbBr2 = 0.0331 moles × 367 g/mol= 12.1 g
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