You want to create 12g of copper to meld into a piece of jewerly. You know that when copper(II) chloride reacts with aluminum, copper is a product. How much aluminum would you (? finished down)

Question

You want to create 12g of copper to meld into a piece of jewerly. You know that when copper(II) chloride reacts with aluminum, copper is a product. How much aluminum would you need to start your reaction with to get 12g of copper? (Write and balance the equation first)

Answer 1

Write a correctly balanced equation:

2Al(s) + 3CuCl₂(aq) ==> 3Cu(s) + 2AlCl₃(aq)

We must work in moles, not grams. So, we will convert 12 g Cu to moles Cu and then see how many moles of Al we need. We will convert that value back to grams of Al to get our answer.

12 g Cu x 1 mole Cu/63.55 g = 0.1889 moles Cu

moles Al needed = 0.1889 moles Cu x 2 moles Al/3 moles Cu = 0.1259 moles Al needed

grams Al needed = 0.1259 moles Al x 26.98 g/mol = 3.396 g = 3.4 g Al needed (to 2 significant figures)

Answer 2

First, to balance the equation, you need to determine the subscripts involved with the molecules containing two elements. Copper(II) chloride would be CuCl₂ since each copper(II) atom has two available valence electrons to bond and each chloride atom needs one more electron to form an octet; to have the two available valence electrons of copper bonded, two chlorides must be present. Aluminum is on its own and doesn't need subscripts.

That takes care of the reactants. Then, to determine the products (apart from copper), you have to see what is left on the reactants side that is not copper. That would be aluminum and chloride, forming aluminum chloride. Aluminum has three valence electrons available to bond and, as stated previously, chloride needs one more electron to form an octet (meaning each chloride will bond with one other atom). For each aluminum atom, three chlorides are needed to form bonds with the available valence electrons. This would give you AlCl₃.

CuCl₂ + Al --> Cu + AlCl₃

To balance the equation, you need to ensure that the number of atoms of each element is equivalent on each side. Notice that the number of chlorides are different on each side. So, we begin by adding a coefficient of 3 in front of CuCl₂ and a coefficient of 2 in front of AlCl₃. Now we need to add coefficients onto Al and Cu to account for the coefficients just placed in front of the molecules. We would need a 2 in front of Al and a 3 in front of Cu.

3CuCl₂ + 2Al --> 3Cu + 2AlCl₃

The next thing you would need to do after balancing the equation is determine how much aluminum you must start with to get 12g of copper.

Convert the 12g of copper to moles. To do this, you must divide 12 by the atomic mass of copper (63.546) to cancel out the grams of copper. The reason you divide by the atomic mass is that you know the atomic mass of 63.546 is also the number of grams present in one mole of copper. 12 divided by 63.546 gives you about 0.18884 moles of copper.

To get to moles of aluminum, you would multiply this by the ratio of moles of aluminum over the moles of copper in the balanced equation. Since the coefficient for aluminum is 2 and the coefficient for copper is 3, you multiply by 2/3 to get 0.125893 moles of aluminum since the moles of copper cancel out. To get the number grams of aluminum, you would multiply the moles of aluminum (0.125893) by the atomic mass of aluminum (26.9815) to get about 3.39679.

To account for sig figs, only use two of the digits of 3.39679 since 12 has two sig figs.

This would mean that you would need to start with 3.4 g of Al to get 12 g of Cu.