Problem 8.131 Part A Acid spills are often neutralized with sodium carbonate or

Question

Problem 8.131

Part A

Acid spills are often neutralized with sodium carbonate or sodium hydrogen carbonate. For the neutralization of acetic acid, the unbalanced equations are
(1) CH3CO2H(l)+Na2CO3(s)CH3CO2Na(aq)+CO2(g)+H2O(l)
(2) CH3CO2H(l)+NaHCO3(s)CH3CO2Na(aq)+CO2(g)+H2O(l)
Balance both equations.

Express your answers as chemical equations separated by a comma. Identify all of the phases in your answer.

2CH3CO2H(l)+Na2CO3(s)2CH3CO2Na(aq)+CO2(g)+H2O(l),CH3CO2H(l)+NaHCO3(s)CH3CO2Na(aq)+CO2(g)+H2O(l)

Part B

How many kilograms of each substance is needed to neutralize a 3.800 gallon spill of pure acetic acid (density = 1.049 g/ml)?

Enter your answers using four significant figures separated by a comma. NOTE: 3.521,5.581 are incorrect

Part C

How much heat in kilojoules is absorbed or liberated in each reaction? See Appendix B for standard heats of formation; Hf = -726.1 kJ/mol for CH3CO2Na(aq).

Enter your answers using three significant figures separated by a comma.

Explanation / Answer

part A

balanced equations

2CH3CO2H(l)+Na2CO3(s) 2CH3CO2Na(aq)+CO2(g)+H2O(l)

2 mol aceticacid = 1 mol Na2co3

CH3CO2H(l)+NaHCO3(s) CH3CO2Na(aq)+CO2(g)+H2O(l)

1 mol aceticacid = 1 mol Na2co3

part B

3.8 gallon aceticacid = 14.385 L

mass of aceticacid = density*volume

     = 14.385*10^3*1.049

       = 15089.9 grams

No of mol of aceticacid = 15089.9/60 = 251.5 mol

from the equations above.

No of mol of Na2CO3 required = 2*251.5 = 503 mol

No of mol of NaHCO3 required = 251.5 mol

part C

need more information.

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