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Solved: Federal regulations set an upper limit of 50 parts

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 115 Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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4
Problem 115

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 HCI. The NH3 reacts with HCI according to: 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 MNaOH to reach the equivalence point, (a) How many grams of NH3 were drawn into the acid solution? (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.) (c) Is this manufacturer in compliance with regulations?

Step-by-Step Solution:
Step 1 of 3

Solution: Step1: a) Volume of NaOH solution used = 13.1 mL = 1000 L = 0.0131 L Molarity of the NaOH solution = 0.0588 M Therefore, number of moles of NaOH = volume of solution in litres x molarity = 0.0131 L x 0.0588 M = 0.00077 mol Since, NaOH is monobasic, so, number of moles of HCl reacted with NaOH = 0.008 mol Volume of HCl solution used = 1.00 x 10 mL = 100 mL = 100 L = 0.100 L 1000 Molarity of the HCl solution = 0.0105 M Therefore, number of moles of HCl = volume of solution in litres x molarity = 0.100 L x 0.0105 M = 0.00105 mol Step2: Now, the reaction between HCl and NH takes plac3in the following manner: HCl(aq) + NH (aq) ------> NH Cl(aq) 3 4 It is clear from the above equation that, 1 mol of HCl requires 1 mol of NH for 3 neutralization. Thus, number of moles of NH drawn fr3 air = number of moles of HCl reacted with NH 3 = total number of moles of HCl - number of moles of HCl reacted with NaOH = 0.00105 - 0.00077 = 0.00028 mol Step3: Now, 1 mol of NH = mola3mass of NH 3 Therefore, 0.00028 mol NH = 0.00028 x molar mass of NH 3 3 = 0.00028 x 17.03 g [molar mass of NH = 17.3 g/mol] = 0.0048 g Thus, 0.0048 g of NH were 3awn into the acid solution. Step4: 23 b) 1 mol of...

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Chapter 4, Problem 115 is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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Solved: Federal regulations set an upper limit of 50 parts

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