Problem 8.131 Part A Acid spills are often neutralized with sodium carbonate or
ID: 520399 • Letter: P
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.
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Part B
How many kilograms of each substance is needed to neutralize a 3.900 gallon spill of pure acetic acid (density = 1.049 g/ml)?
Enter your answers using four significant figures separated by a comma.
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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.
2CH3CO2H(l)+Na2CO3(s)2CH3CO2Na(aq)+CO2(g)+H2O(l),CH3CO2H(l)+NaHCO3(s)CH3CO2Na(aq)+CO2(g)+H2O(l)Explanation / Answer
Part A
(1) 2 CH3CO2H (l)+ Na2CO3 (s) 2 CH3CO2Na (aq)+ CO2 (g) + 2 H2O (l)
(2) CH3CO2H (l) + NaHCO3 (s) CH3CO2Na (aq) + CO2 (g) + H2O (l)
PART (B)
Volume of acetic acid = 3.900 gallon = 3.900 * 3785.4 mL = 14763.06 mL
Mass of acetic acid = density * volume = 1.049 * 14763.06 = 15486.45 g. = 15.5 kg.
Moles of acetic acid = mass / molar mass = 15486.45 / 60.0 = 258.1 mol
2 mol of acetic acid 1 mol of sodium carbonate
then, 184.4 mol of acetic acid = 258.1 / 2 = 129.05 mol of sodium carbonate
Therefore, Mass of sodium carbonate neede = 129.05 * 106 = 13679.3 g. = 13.7 kg.
Similarly,
1 mol of aceticacid = 1 mol of sodium bicarbonate
258.1 ol of acetic acid = 258.1 mol of sodium bicarbonate
Therefore, mass of sodium bicarbonate neede = 258.1 * 84 = 21680.4 g. = 21.68 kg.