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Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)â2AlCl3(s) You are given 26.0 g of aluminum and 31.0 g of chlorine gas.
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 26.0 g of aluminum?
If you had excess aluminum, how many moles of aluminum chloride could be produced from 31.0 g of chlorine gas, Cl2?
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)â2AlCl3(s) You are given 26.0 g of aluminum and 31.0 g of chlorine gas.
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 26.0 g of aluminum?
If you had excess aluminum, how many moles of aluminum chloride could be produced from 31.0 g of chlorine gas, Cl2?
asifasabirLv10
12 Jun 2023
Jamar FerryLv2
17 Dec 2019
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