Calculate the mass of copper needed to react with excess chlorine gas to produce 84g of copper(II) chloride--write a balanced equation first.

Question

I'm having trouble with Stoichiometry, specifically problems like these. I've never had a problem before where the only number given to me is what is produced, so I'm not sure on how to set this up and solve it. Please help.

Answer 1

Finding how much reactant needed to produce a certain amount of product uses the same steps as if you are finding how much product is produced from a certain amount of reactant--you are essentially just working backwards!

To solve this problem you need to first identify:

1. The balanced equation
2. The molar mass of copper (II) chloride
3. The molar mass of copper

Once you have identified these things, you can set it up like a normal stoichiometry problem. Remember, to use the balanced equation, you must be working with moles, so you will use the molar mass to convert your mass of copper (II) chloride into moles first.

Steps to Solve:

1. Write the balanced equation: Cu_(s)+Cl_2(g) --> CuCl_2(s)
2. Identify molar masses of CuCl₂ and Cu
3. CuCl₂ molar mass= 134.44 g/mol
4. Cu molar mass= 35.45 g/mol
5. Convert known mass of CuCl₂ produced into moles:
6. 84g CuCl₂/134.44 g/mol = .62 moles of CuCl₂
7. Use balanced equation to use the stoichiometric ratio of moles of CuCl₂ produced for every mole of Cu reacted. You find this ratio by using the number coefficients in your balanced equation. In this case the ratio is 1:1 so the amount of moles of Cu reacted will be the same as the amount of moles of CuCl₂ produced. However, if this was not the case, this is how you would set up the calculation:
8. .62 mol CuCl₂(1 mol Cu/ 1 mol CuCl₂)= .62 mol Cu
9. Use the molar mass of copper to convert the moles of Cu needed into grams:
10. .62 mol Cu (35.45 g/mol)= 22 g of Cu

Answer: 22g of copper are needed to react with excess chlorine gas to produce 84g of copper (II) chloride.

*note that the answer is to two significant figures since the known value that we started with only has 2 significant figures.

Answer 2

Stoichiometry is all about matching how much you have on both sides of the reaction.

First, let's balance: Cu(s) + Cl₂(g) -> CuCl₂(s)

Any metal is just written by itself. Chlorine, however, is what is called diatomic (H, N, O, F, Cl, Br, and I), which means they always come in a pair if they are not bonded to anything else. On the products side, the copper has a II, which means a charge of +2. Each Cl has a charge of -1 since it is in the far right column. See this video for help with charges. (https://terpconnect.umd.edu/~wbreslyn/chemistry/naming/findingioniccharge.html)

So we need two -1 charges to balance the +2 charge (compounds always are neutral unless they tell you a different charge for it).

Now we have to check if we have the same number of each element on both sides. If not, you would add a number in front of the compounds (this is called a coefficient) to add more of that compound. However, we have 1 Cu and 2 Cl on both sides so we do not need any more of anything.

Now that it is balanced, we can find out how much we need to make 84 g of CuCl₂. When doing so, remember, the coefficients are a ratio of moles. So this means for every 1 mole of Cu, there is 1 mole of Cl₂, and 1 mole of CuCl₂ since all the coefficients are 1.

So we have to convert the 84 g of CuCl₂ to moles first. Using the atomic masses on the periodic table, Cu=63.55 g/mol and Cl=35.45 g/mol. Since we have one Cu and 2 Cl, the total molar mass is 63.55g/mol +2x35.45 g/mol = 134.45 g/mol. This means in one mole of CuCl₂ it would have a mass of 134.45 g. So we can take our 84 g CuCl₂ and divide by the 134.45 g/mol (we divide because grams are what we start with so the 134.45 g must go on the bottom to cancel grams, and the 1 mole goes on top because that is the final unit we want). We get 0.62 moles of CuCl₂.

Since there is a 1:1 ratio for all compounds 0.62 moles of CuCl₂ means 0.62 moles of Cu as well.

Last step. Convert the moles of Cu back to mass. Cu=63.55 g/mol. So 0.62 moles times 63.55 g/mol (this time we multiply because we are starting with moles, so the moles in the molar mass must go on the bottom and the grams on top). Our final mass is 39.7 g of Cu.