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.
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.
General Chemistry - Standalone book (MindTap Course List)
11th Edition
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Chapter5: The Gaseous State
Section: Chapter Questions
Problem 5.119QP
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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.
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