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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 14.0 g of aluminum and 19.0 g of chlorine gas.
1)If you had excess chlorine, how many moles of of aluminum chloride could be produced from 14.0 g of aluminum?
2)If you had excess aluminum, how many moles of aluminum chloride could be produced from 19.0 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 14.0 g of aluminum and 19.0 g of chlorine gas.
1)If you had excess chlorine, how many moles of of aluminum chloride could be produced from 14.0 g of aluminum?
2)If you had excess aluminum, how many moles of aluminum chloride could be produced from 19.0 g of chlorine gas, Cl2?
22 Jun 2023
Jamar FerryLv2
17 Dec 2019
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