Question

For the reaction Cl2(g) + KBr(s) to KCl(s) + Br2(g),how many grams of potassium chloride can be produced from 300 grams of potassium bromide? a.98.7g b.111g c.188g d.451g

Answer 1

See below.

Explanation:

Step 1: Balance the chemical equation.

`Cl_2(g)+2KBr(s)->2KCl(s)+Br_2(g)`

Step 2: Identify the limiting reagent.

This is the compound that will basically run out first, but this problem doesn't give us the amount of `Cl_2` the `KBr` is reacting with, so we assume it is reacting in an excess of `Cl_2` (meaning `KBr` is our limiting reagent)

Step 3: Convert grams to moles using `KBr` molar mass.

`"300 grams KBr"*(("1 mole KBr")/("119.002 grams"))="2.521 moles KBr"` (not using the correct amount of significant figures)

Step 4: Find out how many moles of `KCl` can be formed be 2.521 moles of `KBr` using the molar ratio between the two. We do this because the numbers in our balanced reaction are mole ratios, meaning 2 moles of `KBr` will produce 2 moles of `KCl`.

It is important to note this ratio is NOT a mass ratio (i.e. 300g of `KBr` won't produce 300g of `KCl`

`"2.521 moles KBr"*(("2 moles KCl")/("2 moles KBr"))="2.521 moles KCl"`

Step 5: Convert moles to grams using the molar mass of `KCl`

`"2.521 moles KCl"*(("74.551 grams")/("1 moles KCl"))="187.9 grams KCl"`

Thus, c (188 g) will be the correct answer.

Hope this helps!