Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) You are given 29.0 g of aluminum and 34.0 g of chlorine gas. a) If you had excess chlorine, how many moles of aluminum chloride could be produced from 29.0 g of aluminum? b) If you had excess aluminum, how many moles of aluminum chloride could be produced from 34.0 g of chlorine gas, Cl2?
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) You are given 29.0 g of aluminum and 34.0 g of chlorine gas. a) If you had excess chlorine, how many moles of aluminum chloride could be produced from 29.0 g of aluminum? b) If you had excess aluminum, how many moles of aluminum chloride could be produced from 34.0 g of chlorine gas, Cl2?
Chemistry: An Atoms First Approach
2nd Edition
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Steven S. Zumdahl, Susan A. Zumdahl
Chapter5: Stoichiometry
Section: Chapter Questions
Problem 158CWP: Consider the following unbalanced chemical equation for the combustion of pentane (C5H12):...
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Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:
2Al(s)+3Cl2(g)→2AlCl3(s)
You are given 29.0 g of aluminum and 34.0 g of chlorine gas.
a) If you had excess chlorine, how many moles of aluminum chloride could be produced from 29.0 g of aluminum?
b) If you had excess aluminum, how many moles of aluminum chloride could be produced from 34.0 g of chlorine gas, Cl2?
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