Number 1. I get 33 liters. No idea what I did wrong as I calculate one mole of aluminum present one mole of hydrochloric acid present giving me 1.5 miles of hydrogen plug that in topv= nrt. With p= 1atm. T = 273. Aluminum reacts with hydrochloric acid according to the followingequation:2Al(s) + 6HCI(aq) → 2AIC13(aq) + 3H2(g)1. If a sample of 27.0 g of aluminum metal is added to 333 mL of3.0 M hydrochloric acid, the volume of hydrogen gas pro-duced at standard temperature and pressureis(A) 2.80 L.(B) 5.60 L.(C) 11.2 L.(D) 22.4 L.

Given- Reaction- 2 Al (s) + 6 HCl (aq) →2 AlCl3 (aq) + 3 H2 (g) Mass of Al = 27.0 g Volume of HCl…

Answered: Aluminum reacts with hydrochloric acid…

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter15: Solutions

Section: Chapter Questions

Problem 134AP

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Number 1. I get 33 liters. No idea what I did wrong as I calculate one mole of aluminum present one mole of hydrochloric acid present giving me 1.5 miles of hydrogen plug that in to

pv= nrt. With p= 1atm. T = 273.

Aluminum reacts with hydrochloric acid according to the following
equation:
2Al(s) + 6HCI(aq) → 2AIC13(aq) + 3H2(g)
1. If a sample of 27.0 g of aluminum metal is added to 333 mL of
3.0 M hydrochloric acid, the volume of hydrogen gas pro-
duced at standard temperature and pressure
is
(A) 2.80 L.
(B) 5.60 L.
(C) 11.2 L.
(D) 22.4 L.

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