Nitric acid solution spilled. Spill was neutralized with solid sodium carbonate,

Question

Nitric acid solution spilled. Spill was neutralized with solid sodium carbonate, also releasing non-flammable gas a) Write the balanced neutralization equation b) approx 2.00 times 10^4 gallons of nitric acid solution spilled calculate the volume in liters of acid solution spilled. c) assume that the acid was an aqueous solution containing 70.0% nitric acid by mass with solution density of 1.42 g/mL Find the mass in grams of pure nitric acid spilled. d) How much sodium carbonate in kg was required for complete neutralization of the spilled nitrification?

Explanation / Answer

Ans. 8a. Na2CO3(s) + HNO3(aq) -----------> NaNO3(aq) + H2O(l) + CO2(g)

Stoichiometry: 1 mol solid carbonate (Na2CO3) reacts with 1 mol nitric acid (HNO3) for neutralization giving 1 mol sodium nitrate (NaNO3), 1 mol water and 1 mol carbon dioxide (CO2) as non-flammable gas.

Ans. 8b. Volume of HNO3 spillage = 2.00 x 10­4 gallons

                                                = (2.00 x 10­4) x 1 gallons

                                                = (2.00 x 10­4) x (3.78541 L)        ; [1 gallon = 3.78541 L ]

                                                = 75708.236 L

Ans. 8c. Given, density of HNO3 solution = 1.42 g /mL

                                                            = 1.42 kg/ L

Total mass of HNO3 spillage = Volume of HNO3 spillage x density of HNO3

                                    = 75708.236 L x (1.42 kg/ L)

                                    = 107505.69512 kg

Given, % of pure HNO3 is spillage = 70%

So, Mass of pure HNO3 in spillage = 70% of 107505.69512 kg

                                                = 75253.986584 kg

Ans. D. Moles of HNO3 spillage = Mass of pure HNO3 spilled / molar mass of HNO3

                                                                = 75253.986584 kg x (63.012 g mol-1)

                                                = 75253986.584 g x (63.012 g mol-1)       ; [1 kg = 1000 g]

                                                = 1194280.24160 moles

From stoichiometry, 1 mol HNO3 is neutralized by 1 mol sodium carbonate.

So, moles of sodium carbonate required for complete neutralization = 1194280.24160 moles     

Mass of Na2CO3 required = moles of Na2CO3 x molar mass of Na2CO3

                                    = 1194280.24160 moles x (105.99 g mol-1)

                                    = 126581762.80768996381641 gram

                                    = 126581.76 kg

Thus, the required mass of Na2CO3 = 126581.76 kg = 1.27 x 105 kg

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.