Federal regulations set an upper limit of 50 parts per million (ppm) of NH3 in the air in a work environment [that is, 50 molecules of NH3 for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00 * 102 mL of 0.0105 M HCl. The NH3 reacts with HCl according to: NH3(aq) + HCl(aq) → NH4Cl(aq). After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. Is this manufacturer in compliance with regulations?

Verified step by step guidance

Step 1: Calculate the initial moles of HCl in the solution. Use the formula \( \text{moles} = \text{volume} \times \text{molarity} \). The initial volume of HCl is 100 mL (or 0.100 L) and the molarity is 0.0105 M.

Step 2: Determine the moles of NaOH used in the titration. Use the formula \( \text{moles} = \text{volume} \times \text{molarity} \). The volume of NaOH used is 13.1 mL (or 0.0131 L) and the molarity is 0.0588 M.

Step 3: Calculate the moles of HCl that reacted with NH3. Since NaOH neutralizes the remaining HCl, subtract the moles of NaOH from the initial moles of HCl to find the moles of HCl that reacted with NH3.

Step 4: Determine the total volume of air passed through the solution. The air was drawn at a rate of 10.0 L/min for 10.0 minutes, resulting in a total volume of 100 L.

Step 5: Calculate the concentration of NH3 in the air in ppm. Use the moles of NH3 (equal to the moles of HCl that reacted) and the total volume of air to find the concentration in ppm. Compare this value to the regulatory limit of 50 ppm to determine compliance.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molarity and Concentration

Molarity (M) is a measure of concentration defined as the number of moles of solute per liter of solution. In this scenario, the concentration of HCl is given as 0.0105 M, indicating that there are 0.0105 moles of HCl in every liter of the solution. Understanding molarity is crucial for calculating the amount of NH3 that can react with the HCl in the solution.

Stoichiometry of Chemical Reactions

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced equations. The reaction between NH3 and HCl produces NH4Cl in a 1:1 molar ratio. This concept is essential for determining how much NH3 was present in the air sample by relating the amount of HCl that reacted to the amount of NH3 produced.

Titration and Equivalence Point

Titration is a quantitative analytical method used to determine the concentration of a solute in a solution. The equivalence point is reached when the amount of titrant added is stoichiometrically equivalent to the amount of substance being analyzed. In this case, the volume of NaOH used in the titration indicates how much HCl remained unreacted, allowing for the calculation of the amount of NH3 that was absorbed from the air.

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Potassium superoxide, KO₂, is often used in oxygen masks (such as those used by firefighters) because KO₂ reacts with CO₂ to release molecular oxygen. Experiments indicate that 2 mol of KO₂(s) react with each mole of CO₂(g). (b) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced?

Textbook Question

Federal regulations set an upper limit of 50 parts per million (ppm) of NH₃ in the air in a work environment [that is, 50 molecules of NH₃(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00⨉10² mL of 0.0105 M HCl. The NH₃ reacts with HCl according to: NH₃₍aq) + HCl(aq) → NH₄Cl(aq). After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. (a) How many grams of NH₃ were drawn into the acid solution?

Textbook Question

Federal regulations set an upper limit of 50 parts per million (ppm) of NH₃ in the air in a work environment [that is, 50 molecules of NH₃(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00⨉10² mL of 0.0105 M HCl. The NH₃ reacts with HCl according to: NH₃₍aq) + HCl(aq) → NH₄Cl(aq). After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. (b) How many ppm of NH₃ were in the air? (Air has a density of 1.20 g/L and an average molar mass of 29.0 g/mol under the conditions of the experiment.)

Textbook Question

Chlorine dioxide gas 1ClO22 is used as a commercial bleachingagent. It bleaches materials by oxidizing them. In thecourse of these reactions, the ClO2 is itself reduced. (b) Why do you think that ClO2 is reduced so readily?