Question

Question #67db3

Answer 1

you would get about 35.8 grams of `Br_2` from this reaction.

Explanation:

Let's begin with balancing our equation and figuring how much we have of everything.

The reaction between Calcium Bromide and Chlorine will be a single replacement reaction. Therefore, we know the reaction will look a little like this:

`CaBr_2 + Cl_2 -> CaCl_2 + Br_2`

Luckily we don't need to try and balance this equation, as it is already balanced for us. Next, let's find out how many moles of each reactant we have to find out the limiting reactant:

`66.8 " g " CaBr_2 xx (1 " mol")/(199.9 " g") = .334 " mol " CaBr_2`

The Chlorine situation is a little more tricky, as we have to use the ideal gas law to solve for n, or moles of Chlorine:

`PV = nRT` where
P = Pressure, in atm
V = Volume, in Liters,
n = moles of gas
R = The universal gas constant, 0.08206 `(L*atm)/(mol*K)`
T = Absolute temperature, in Kelvin

When gases are at STP (Standard temperature & pressure), the temperature is 273 K and the pressure is 1 atm. Now we can plug in the given value of V and solve for n.

`PV = nRT`

`1 xx 9.96 = n xx .08206 xx 273`

`n = 0.407` moles of Cl.

So to review, we have determined the balanced equation for the reaction between `CaBr_2` and `Cl_2`, and have found out how many moles we have of each reactant. Moving on.

As hinted by the reaction `CaBr_2 + Cl_2 -> CaCl_2 + Br_2`, the ratio of `CaBr_2` to `Cl_2` is 1:1.

For every 1 mole of `CaBr_2` you will need 1 mole of `Cl_2`. Using this logic, it is easy to see that `CaBr_2` will be the limiting reactant as you have more moles of `CaBr_2` than `Cl_2`.

Now that we have determined `CaBr_2` to be our limiting reactant, we can now convert it to the desired product by using the balanced chemical equation.

`.334 " mol" CaBr_2 xx (1" mol " Br_2)/(1" mol " CaBr_2) = .334 " moles of "Br_2" formed."`

We are almost there! However, In the prompt it says that the process is only 67.0% efficient, so we would only get 67% of .334 which is equal to .224 moles of `Br_2`.

Finally, converting this to the mass of `Br_2`,

`.224 " mol " Br_2 xx (159.8 " g ")/(1 " mol ") = 35.8 " grams of "Br_2" formed."`