Challenge Problems HW 9, Week 11 Due week of 4/2 Bill Nye\'s daughter, Pink Nye,
ID: 1033949 • Letter: C
Question
Challenge Problems HW 9, Week 11 Due week of 4/2 Bill Nye's daughter, Pink Nye, gets carried away in lab and mixes several salt solutions together First, she mixed 10 mL of 1.5 M Ba(OH)2 with 15 mL of 2.0 M NH4Cl, which react according to the following reaction: 1. 2NHACI(aq) + Ba(OH)2(aq) 2H20(I) + 2NH3(aq) + Ba2+ (aq) + Cl-(aq) Next, she added 7.90 mL of 1.5 M Pb(NO3)2 and a white precipitate formed. She centrifuged (you can ask your TA what this means) the solution down so that the precipitate was collected at the bottom of her flask. Lastly, she added 23.5 mL of 0.50 M Na2COs and more white solid formed. assume no other reactions happen other than the given equation and any p reactions* a. List all precipitates and their theoretical yields(in grams) b. List all aqueous components and give their final concentrations (in M)Explanation / Answer
a) Hi Dear Friend, answers are in BOLD
Moles Ba(OH)2 = 10 mL x 10-3 L x 1.50 M = 0.015 mol
Moles NH4Cl = 15 mL x 10-3 L x 2.0 M = 0.03 mol
According to given balanced chemical reaction equation 2 mol NH4Cl reacts wit 1 mol Ba(OH)2, hence 0.03 mol NH4Cl requires 0.015 mol Ba(OH)2 and forms 2 mol H2O, 2 mol NH3 and 1 mol BaCl2. This means 0.03 mol H2O, 0.03 mol NH3 and 0.015 mol BaCl2 formed
Now she added 7.9 mL of 1.5M Pb(NO3)2
BaCl2 (aq) + Pb(NO3)2 --> Ba(NO3)2(aq) + PbCl2(ppt)
moles Pb(NO3)2 = 7.9 mL x 10-3 L x 1.5M = 0.01185 mol
0.01185 mol Pb(NO3)2 reacts with equal number of moles of BaCl2 according to above equation. So we have 0.01185 moles PbCl2 precipitate (ppt) and 0.01185 moles Ba(NO3)2 also in solution.
Theoretical yield of PbCl2 precipitate = moles PbCl2 x molar mass of PbCl2 = 0.01185 mol x 278.1 g/mol
Theoretical yield of PbCl2 precipitate = 3.295485 grams
Now she added 23.5 mL of 0.50M Na2CO3
PbCl2 is already precipitated, hence now decant solution contains Ba(NO3)2 it reacts with Na2CO3
Na2CO3(aq) + Ba(NO3)2(aq) --> BaCO3 (ppt) + 2NaNO3(aq)
moles Na2CO3 = 23.5 mL x 10-3 L x 0.50M = 0.01175 mol
0.01175 mol Na2CO3 reacts with 0.01175 mol Ba(NO3)2 only and forms 0.01175 mol BaCO3 precipitate.
Theoretical yield of BaCO3 precipitate = moles BaCO3 x molar mass of BaCO3 = 0.01175 mol x 197.34 g/mol
Theoretical yield of BaCO3 precipitate = 2.318745 grams
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b) Total volume before addition of Na2CO3 = 10 mL + 15 mL + 7.9 mL = 32.9 mL = 0.0329 L
H2O concentration = moles H2O / solution volume in liters = 0.03 mol / 0.0329 L = 0.912 M
NH3 concentration = moles NH3 / solution volume in liters = 0.03 mol / 0.0329 L = 0.912 M
only 0.01175 moles of Ba(NO3)2 reacted out of 0.01185 moles, hence remaining Ba(NO3)2 moles in solution = 0.0001 mol only.
Ba(NO3)2 concentration = moles Ba(NO3)2 / solution volume in liters = 0.0001 mol / 0.0329 L = 0.003M
0.01175 mol Na2CO3 gives 2 x 0.01175 mol = 0.0235 mol NaNO3
New volume after addition of Na2CO3 = 32.9 mL + 23.5 mL = 56.4 mL = 0.0564 L
NaNO3 concentration = moles NaNO3 / solution volume in liters = 0.0235 mol / 0.0564 L = 0.42M
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