Asked by chris
From the Periodic Table, 1 mole of aluminum weighs 27.0 g.,
Therefore;
If 54.0 g of aluminum are placed into the beaker to react with copper(II) chloride, how many grams of the metal product will be produced?
Therefore;
If 54.0 g of aluminum are placed into the beaker to react with copper(II) chloride, how many grams of the metal product will be produced?
Answers
Answered by
Steve
If the reaction is
3CuCl<sub>2</sub> + 2Al = 2AlCl<sub>3</sub> + 3Cu
looks like you have 3 moles of copper for every 2 moles of aluminum.
You have 2 moles of aluminum.
3CuCl<sub>2</sub> + 2Al = 2AlCl<sub>3</sub> + 3Cu
looks like you have 3 moles of copper for every 2 moles of aluminum.
You have 2 moles of aluminum.
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