Federal regulations set an upper limit of 50 parts per million (ppm) of NH 3 in the air in a work environment [that is, 50 molecules of NH 3 (g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00⨉10 2 mL of 0.0105 M HCl. The NH 3 reacts with HCl according to: NH 3( aq) + HCl(aq) → NH 4 Cl(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 NH 3 were drawn into the acid solution?

Hi everyone here we have a question that says the who guidelines recommend that the amount of residual chlorine in drinking water should not be higher than five parts per million. The amount of corn can be determined by backed iteration with potassium iodide and that equation is chlorine nucleus plus two iodine. Aquarius forms to chloride Aquarius plus iodine Aquarius. The iodine formed as a result of the reaction above is then tie traded with the solution of sodium theo sulfate to sodium theo sulfate plus iodine. Aquarius forms to sodium iodide Aquarius plus sodium teacher Thean eight A 350.0 mL. Water sample was added with excess solid sodium iodide and the resulting solution was then titrate ID against a 2.25 times 10 to the negative three molar sodium theo sulfate solution. The solution required 14.5 ml of sodium theo sulfate to reach the equivalence point. How many grams of chlorine are there in the water sample? So we're going to start off with our 14.5 ml of sodium Theosophy eight And we're going to multiply that by 10 to the -3. Leaders over one millimeter And we're going to multiply that by 2. Times 10 to the -3 moles of sodium theo sulfate over one leader because polarity is moles over leaders And we're going to multiply that by one mole of I diane over two moles of sodium theo sulfate And we're going to multiply that by one mole of chlorine over one mole of iodine? And we're going to multiply that by chlorine molar mass, which is 70 .90 g of chlorine Over one mole of coin. So our middle leaders of sodium theo sulfate are canceling out. Our leaders are canceling out are moles of sodium theo sulfate are canceling out. Our moles of iodine are canceling out and our moles of chloride are canceling out. And that gives us 0. 1157. If we want to put that into scientific notation, that would be 1.16 times 10 to the negative three grams of chlorine. And that is our final answer. Thank you for watching. Bye.

Ch.4 - Reactions in Aqueous Solution

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Brown 14th Edition

Ch.4 - Reactions in Aqueous Solution

Problem 115a

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Chapter 4, Problem 115a

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?

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Calculate the total volume of air drawn through the solution: \(10.0 \text{ L/min} \times 10.0 \text{ min} = 100.0 \text{ L}\).

Determine the initial moles of HCl in the solution: \(0.0105 \text{ M} \times 0.100 \text{ L} = 0.00105 \text{ mol HCl}\).

Calculate the moles of NaOH used in the titration: \(0.0588 \text{ M} \times 0.0131 \text{ L} = 0.00076968 \text{ mol NaOH}\).

Determine the moles of HCl that reacted with NH\(_3\) by subtracting the moles of NaOH from the initial moles of HCl: \(0.00105 \text{ mol} - 0.00076968 \text{ mol} = 0.00028032 \text{ mol HCl reacted}\).

Convert the moles of NH\(_3\) to grams using the molar mass of NH\(_3\) (17.03 g/mol): \(0.00028032 \text{ mol} \times 17.03 \text{ g/mol}\).

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Key Concepts

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

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It allows us to determine the amount of substances consumed or produced in a reaction. In this case, understanding the stoichiometric relationship between NH3 and HCl is crucial for calculating how much NH3 reacted with the acid.

Molarity and Dilution

Molarity (M) is a measure of concentration defined as moles of solute per liter of solution. In this problem, the concentration of HCl is given, and we need to use it to find the number of moles of HCl that reacted. Additionally, the dilution of HCl after the reaction with NH3 will be important for determining how much acid remains, which can be used to find the amount of NH3 that was absorbed.

Titration

Titration is a laboratory technique used to determine the concentration of a solute in a solution by reacting it with a reagent of known concentration. In this scenario, the remaining HCl after the reaction with NH3 is titrated with NaOH to find out how much HCl was left, which helps in calculating the amount of NH3 that reacted. The equivalence point in titration indicates that the amount of titrant added is stoichiometrically equivalent to the amount of substance being analyzed.

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Acid-Base Titration

Related Practice

The arsenic in a 1.22-g sample of a pesticide was converted to AsO₄^3- by suitable chemical treatment. It was then titrated using Ag⁺ to form Ag₃AsO₄ as a precipitate. (c) If it took 25.0 mL of 0.102 M Ag⁺to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion (ppb) arsenic. If this arsenic is present as arsenate, AsO4^3-, what mass of sodium arsenate would be present in a 1.00-L sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as ppb = (g solute / g solution) × 10⁹.

Textbook Question

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?

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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.)

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Textbook Question

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

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