Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:2Al(s) + 3Cl_2(g) -gt; 2AlCl_3(s)What is the maximum mass of aluminum chloride that can be formed when reacting 32.0 g of aluminum with 37.0 g of chlorine?Express your answer numerically in grams.Someone please explain and help thank you!
you have to go down to the molar mass which of aluminum would by 26.982 g/mol and in chlorine would be 35.453 g/mol sosince asuminum and chlorine react as AlCl3 than you have a third as much aluminum and if you devide 37 by 35.453 you get 104.3635.% multiply this by 26.982 to get the same amount of atoms of aluminum as of chlorine this would be 28.15936.g, but you need a third as many so it would be that devided by 3 which is 9.386.g than add the full weight of chlorine so 9.386 +37 which is 46.386455307026203.g of aluminum chloride just in case your teacher doesn't like rounding.
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