If 1.35 g of aluminum is mixed with 6.0 moles of copper (II) chloride, what mass of copper would be produced?

1 Answer
Aug 22, 2017

Answer:

Since aluminum is the limiting reactant, so the reaction will produce #"4.77 g Cu"#.

Explanation:

Start with a balanced equation.

#"2Al + 3CuCl"_2##rarr##"2AlCl"_3 + "3Cu"#

Convert #"1.35 g Al"# to moles by dividing by its molar mass, #"26.982 g/mol"#. Since molar mass is a fraction, g/mol, you can divide by multiplying the given mass by the inverse of the molar mass, mol/g .

#1.35color(red)cancel(color(black)("g Al"))xx(1"mol Al")/(26.982color(red)cancel(color(black)("g Al")))="0.0500 mol Al"#

You need to determine the limiting reactant (or limiting reagent). You need to determine which element or compound produces the least amount of #"Cu"#.

The process involves multiplying the mol #"Al"# and #"CuCl"_2"# by the mole ratio between each of them and #"Cu"#, and then multiply by the molar mass of #"Cu"#, #"63.546 g/mol"#.

Possible Mass of Cu from Al

#0.0500color(red)cancel(color(black)("mol Al"))xx(3color(red)cancel(color(black)("mol Cu")))/(2color(red)cancel(color(black)("mol Al")))xx(53.546"Cu")/(1color(red)cancel(color(black)("mol Cu")))="4.77 g Cu"#

Possible Mass of Cu from #"CuCl"_2#

#6.0color(red)cancel(color(black)("g CuCl"_2))xx(3color(red)cancel(color(black)("mol Cu")))/(3color(red)cancel(color(black)("mol CuCl"_2)))xx(63.546"g Cu")/(1color(red)cancel(color(black)("mol Cu")))="380 g Cu"#

Under the conditions given, Al is the limiting reactant because it will produce only #"4.77 g Cu"#, whereas, if there were enough Al present, the #"CuCl"_2"# would produce #"380 g Cu"#. However, under the conditions given, the Al would run out before the reaction could produce #"380 g Cu"#. The #"CuCl"_2# is the reactant in excess.