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4c) is the tricky one must have confused myself and got it wrong but I got A) and B) correct as I will attach my work in a picture below as well thank you.

4c) is the tricky one must have confused myself and got it wrong but I got A) and B) correct as I will attach my work in a picture below as well thank you.

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter4: Chemical Reactions
Section: Chapter Questions
Problem 4.70QP
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4c) is the tricky one must have confused myself and got it wrong but I got A) and B) correct as I will attach my work in a picture below as well thank you.

Chem 111 > Exams 3) A 50.0 mL of 0.250 M BaCl2 solution were added to 50.0 mL of 0.160M Al2(SO4)3 solution. The reaction equation: 3 ВаСlz (aq) + | Al2(SO4)3 (aq) З BaSO4 (s) I nol + 2 AICI3 (aq) (a) What is the limiting reactant? M x V (0.125) o4 Ale (soys n of Ba Ciį = 2 0.00 4z nols :(6.250m)(50.0 × 10L) 3 nol Ba Clz Eccres ot Al, csouls = %3D 0.008 nols - 0.0042 nols = 0.0125 nola 0. 0038 ols not Al, lsoy)s= M«v Ba Cle is lini ting reactan + = (0.160 M)( s0.0x 6)= 0. 008 mols Answer: (b) Calculate the mass of BaSO4 (233.4g/mol). n of Basoy - 3 nols Basoy(0.125) = 0.12s mols Basoy 3 nols Bacle MM Bu soy- 233.4gInol - (0.0125 yaet ) (233.49/mat) = 2.9!75 9 Mass of Ba soy = 2.929 Ba soy Mas Ba Soy = n Macs of Ba Soy =? (c) What are the molar concentrations (molarities) of the remaining ions in the solution? Answer: Al ces Alelsouls = 0.0038 M mols En cecs nols o4 Alcl 3 = 2 nols of Alcls M nols (.0125 nok) 3 nok of Ba clz 0.0038 0.1 = 0. 00% 3 nols of Al ces, nols k to tal V Total v= 50.OmL + So.0wt of Alel so,)s = COcentra t ion of Allls = Concentration 0.0 38 M nols Bu S0y = O. 012s ols - Jee o ne = 0.1 L conc entra tion nols O. 0I2s nols 0.1 concentration of Ba Soy - 0.125 M Concentrafion of Alles = 0.0083nol 0.1 L Concentrafion of Alles = 0.08 3 M %3D Answer: 1) 2) 3) 99+ 3:43 PM ») 11/3/2020
Transcribed Image Text:Chem 111 > Exams 3) A 50.0 mL of 0.250 M BaCl2 solution were added to 50.0 mL of 0.160M Al2(SO4)3 solution. The reaction equation: 3 ВаСlz (aq) + | Al2(SO4)3 (aq) З BaSO4 (s) I nol + 2 AICI3 (aq) (a) What is the limiting reactant? M x V (0.125) o4 Ale (soys n of Ba Ciį = 2 0.00 4z nols :(6.250m)(50.0 × 10L) 3 nol Ba Clz Eccres ot Al, csouls = %3D 0.008 nols - 0.0042 nols = 0.0125 nola 0. 0038 ols not Al, lsoy)s= M«v Ba Cle is lini ting reactan + = (0.160 M)( s0.0x 6)= 0. 008 mols Answer: (b) Calculate the mass of BaSO4 (233.4g/mol). n of Basoy - 3 nols Basoy(0.125) = 0.12s mols Basoy 3 nols Bacle MM Bu soy- 233.4gInol - (0.0125 yaet ) (233.49/mat) = 2.9!75 9 Mass of Ba soy = 2.929 Ba soy Mas Ba Soy = n Macs of Ba Soy =? (c) What are the molar concentrations (molarities) of the remaining ions in the solution? Answer: Al ces Alelsouls = 0.0038 M mols En cecs nols o4 Alcl 3 = 2 nols of Alcls M nols (.0125 nok) 3 nok of Ba clz 0.0038 0.1 = 0. 00% 3 nols of Al ces, nols k to tal V Total v= 50.OmL + So.0wt of Alel so,)s = COcentra t ion of Allls = Concentration 0.0 38 M nols Bu S0y = O. 012s ols - Jee o ne = 0.1 L conc entra tion nols O. 0I2s nols 0.1 concentration of Ba Soy - 0.125 M Concentrafion of Alles = 0.0083nol 0.1 L Concentrafion of Alles = 0.08 3 M %3D Answer: 1) 2) 3) 99+ 3:43 PM ») 11/3/2020
3) A 50.0 mL of 0.250 M BaCl2 solution were added to 50.0 mL of 0.160M Al2(SO4)3 solution. The reaction equation: 3 BaCl2 (aq) + Al2(SO4); (aq) 3 BaSO4 (s) + 2 AIC1; (aq) (a) What is the limiting reactant? Answer: (b) Calculate the mass of BaSO4 (233.4g/mol). Answer: (c) What are the molar concentrations (molarities) of the remaining ions in the solution?
Transcribed Image Text:3) A 50.0 mL of 0.250 M BaCl2 solution were added to 50.0 mL of 0.160M Al2(SO4)3 solution. The reaction equation: 3 BaCl2 (aq) + Al2(SO4); (aq) 3 BaSO4 (s) + 2 AIC1; (aq) (a) What is the limiting reactant? Answer: (b) Calculate the mass of BaSO4 (233.4g/mol). Answer: (c) What are the molar concentrations (molarities) of the remaining ions in the solution?
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