1.Given the reaction: 2 Al (s) + 6 HCl (aq) --> 2 AlCl3 (aq) + 3 H2 (g)A strip of aluminum of unknown mass is reacted with excesshydrochloric acid. When the hydrogen gas is collected overwater, it is done so at a temperature of 22.0 oC and fills aflask of volume 43.8 mL. The pressure inside thecollection tube is 750.3 mmHg and the water vaporpressure at 22.0 oC is 19.83 torr. Determine the mass of the aluminum strip in milligrams (mg).2.What is the concentration of potassium nitrate (in molarity) at saturation if the mass solubility at 50 oC is 88 g KNO3 per 100 g water?3.What is the partial pressure (in mmHg) of a 39.2% by mass a solution ofpropylene glycol (C3H8O2) dissolved in water at 96.0 oC? Thevapor pressure of pure water at 96.0 oC is 657.6 mmHg. Assume the propylene glycol solution is an ideal solution.

“Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved, then please specify the question number or post only that question.”1) The given balanced chemical reaction is shown below: 2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g) According to the question, when the hydrogen gas is collected over water, the temperature becomes 22.0 oC The volume of hydrogen gas that fills a flask = 43.8 mL = 0.0438 L The pressure inside the collection tube = 750.3 mmHg = 0.987 atm The water vapor pressure at 22.0 oC = 19.83 torr = 0.0260 atm Ptotal=PH2+PH2O0.987 atm =PH2+0.0260 atm PH2=0.987 atm -0.0260 atm PH2=0.961 atm According to the ideal gas equation the number of moles of H2 gas is calculated as follows: PV=nRT n=PVRT =0.961 atm×0.0438 L0.0821 Latm K-1mol-1×295 K =1.74×10-3mol Therefore, the number of moles of H2 gas produced is 1.74 x 10-3 mol.As per the balanced chemical equation, 3.00 mol of H2 produced by 2.00 mol of Al. Therefore, the number of moles Al required to produce 1.74 x 10-3 mol of H2 are calculated as follows: nAl=23×1.74×10-3mol =1.16×10-3mol Molar mass of Al = 27 g/mol According to the mole concept the mass of Al is calculated as follows: mass of Al=n×Mw =1.16×10-3mol×27 g/mol =0.03132 g×1000 mg1 g =31.32 mg Thus, the mass of the aluminum strip in milligrams (mg) is 31.32.

Answered: 1. Given the reaction: 2 Al…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Given the reaction:

2 Al (s) + 6 HCl (aq) --> 2 AlCl₃ (aq) + 3 H₂ (g)

A strip of aluminum of unknown mass is reacted with excess

hydrochloric acid. When the hydrogen gas is collected over

water, it is done so at a temperature of 22.0 ^oC and fills a

flask of volume 43.8 mL. The pressure inside the

collection tube is 750.3 mmHg and the water vapor

pressure at 22.0 ^oC is 19.83 torr.

Determine the mass of the aluminum strip in milligrams (mg).

What is the concentration of potassium nitrate (in molarity) at saturation if the mass solubility at 50 ^oC is 88 g KNO₃ per 100 g water?

What is the partial pressure (in mmHg) of a 39.2% by mass a solution of

propylene glycol (C₃H₈O₂) dissolved in water at 96.0 ^oC? The

vapor pressure of pure water at 96.0 ^oC is 657.6 mmHg.

Assume the propylene glycol solution is an ideal solution.

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