Federal regulations set an upper limit of 50 parts per million \((\mathrm{ppm})\) of \(\mathrm{NH}_{3}\) in the air in a work environment \([\mathrm{that}\) is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{mL}\) of 0.0105 \(\mathrm{M}\) HCl. The \(\mathrm{NH}_{3}\) reacts with HCl according to: $$\mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)$$ After drawing air through the acid solution for 10.0 min at a rate of 10.0 \(\mathrm{L} / \mathrm{min}\) , the acid was titrated. The remaining acid needed 13.1 \(\mathrm{mL}\) of 0.0588 \(\mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of 1.20 g/L and an average molar mass of 29.0 \(\mathrm{g} / \mathrm{mol}\) under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?

Step by step solution

01

Calculate moles of HCl initially in the solution

To find the moles of HCl initially in the solution, we use the formula: moles = concentration × volume. The concentration of HCl is 0.0105 M and the volume is \(1.00 \times 10^2\) mL. Make sure to convert mL into L (1 mL = 0.001 L) to keep the units consistent. Calculating the moles of HCl initially in the solution: moles of HCl = (0.0105 M) × (1.00 × 10² mL × 0.001 L/mL) = 0.0105 mol

02

Calculate moles of NaOH at the equivalence point

The equivalence point occurs when the moles of acid and base are equal. In this case, it's moles of HCl and NaOH. Use the same formula as earlier, moles = concentration × volume, where the concentration of NaOH is 0.0588 M and the volume needed is 13.1 mL. Calculating the moles of NaOH at the equivalence point: moles of NaOH = (0.0588 M) × (13.1 mL × 0.001 L/mL) = 0.00076828 mol

03

Calculate moles of NH₃ that reacted with HCl

Knowing the moles of HCl initially in the solution and the moles of NaOH at the equivalence point, we can now find the moles of NH₃ that reacted with HCl. From the balanced equation, the mole ratio of NH₃:HCl is 1:1. calculating moles of NH₃ that reacted with HCl: moles of NH₃ = moles of HCl_initial - moles of NaOH_equiv_point moles of NH₃ = 0.0105 mol - 0.00076828 mol = 0.00973172 mol

04

Calculate grams of NH₃ that reacted with HCl

Now convert the moles of NH₃ into grams using the molar mass of NH₃ (17.03 g/mol). calculating grams of NH₃ that reacted with HCl: grams of NH₃ = moles of NH₃ × molar mass of NH₃ grams of NH₃ = 0.00973172 mol × 17.03 g/mol ≈ 0.1656 g This is the answer to part (a), there were 0.1656 grams of NH₃ drawn into the acid solution.

05

Calculate the number of moles of air drawn through the solution

Use the air density (1.20 g/L) and molar mass of air (29.0 g/mol) to find the moles of air drawn through the solution. The volume of air drawn through the solution is given as 10.0 L/min for 10.0 min, which is a total of 100 L of air. Calculating the mass of air drawn: mass of air = (density of air) × (volume of air) = (1.20 g/L) × (100 L) = 120 g Calculating the number of moles of air drawn: moles of air = mass of air / molar mass_air = 120 g / 29.0 g/mol ≈ 4.14 mol

06

Calculate the concentration of NH₃ in ppm

Now, we can find the concentration of NH₃ in parts per million (ppm) using the moles of NH₃ and moles of air. Calculating the concentration of NH₃ in ppm: ppm of NH₃ = (moles of NH₃ / moles of air) × 10^6 = (0.00973172 mol / 4.14 mol) × 10^6 ≈ 2350 ppm This is the answer to part (b), the concentration of NH₃ in the air was 2350 ppm.

07

Check compliance with federal regulations

The federal regulations state that the maximum allowable concentration of NH₃ in the air is 50 ppm. We calculated the concentration to be 2350 ppm, which is much higher than the limit: Is the manufacturer in compliance? No, the concentration of NH₃ in the air (2350 ppm) is higher than the federal regulations limit (50 ppm). This is the answer to part (c), the manufacturer is not in compliance with federal regulations.

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