when chlorine gas is bubbled into a solution of sodium bromide, the sodium bromide reacts to give bromine a red brown liquid and sodium chlorine. A solution was made by dissolving 20.0g of sodium bromide in 100.0 g of water. After passing through the solution, it contained 16.0 g of bromine and 11.7 g of sodium chloride. How many grams of chlorine reacted?

Cl2 + 2NaBr ==> Br2 + 2NaCl

.x.....20g......16.0g...11.7g

The easy way to work the problem is to evoke the Law of Conservation of Mass. So x + 20 = 16 + 11.7 and solve for x. I think this is what the problem is illustrating; however, there is something wrong with the answer.

If we do it by stoichiometry, we can check to see if the 16 and 11.7 add up right.
mols Br2 = g/molar mass = 16.0/159.8 = about 0.1
mols NaCl is 2x that = 0.2
g NaCl = mols x molar mass = 0.2 x 58.44 = 11.7 and that checks out.

To check the Cl2 answer, mols Cl2 reacted = same as mols Br2 formed = 0.1.
Then g Cl2 = 0.1 mol x 70.9 = 7.09 which is not the same as the x above. (7.09 vs 7.7).

So which is the right answer? I would go with the last one since that agrees with the stoichiometry calculations. My guess is that the author of the problem overlooked this detail or you made a typo and the 20.0 for NaBr should have been 20.6.

Student

Ah, chemistry! The theater of elements and reactions, where bromine steals the show with its red-brown appearance. Let's dive into your question!

To find out how many grams of chlorine reacted, we need to analyze the given information. We started with 20.0g of sodium bromide, and after the reaction, we ended up with 16.0g of bromine and 11.7g of sodium chloride.

Since the bromide in sodium bromide reacted to form bromine, we know that the difference between the initial and final bromide is the bromine produced. Thus, the bromine produced is:

Initial bromide - Final bromide = 20.0g - 16.0g = 4.0g

Now, since sodium chloride is formed by the reaction of chlorine with sodium bromide, we need to determine the amount of chlorine used. The amount of sodium chloride formed is:

Final sodium chloride = 11.7g

The ratio of sodium bromide to sodium chloride is 1:1, meaning that the amount of chlorine used would be equal to the amount of sodium chloride formed, which is 11.7g.

So, the answer is that 11.7 grams of chlorine reacted in this reaction.

To determine the amount of chlorine gas that reacted, we need to consider the stoichiometry of the reaction. The balanced chemical equation for the reaction between chlorine gas (Cl₂) and sodium bromide (NaBr) is as follows:

Cl₂ + 2NaBr -> 2NaCl + Br₂

From the equation, we can see that one mole of chlorine gas reacts with two moles of sodium bromide to produce two moles of sodium chloride and one mole of bromine.

First, let's calculate the number of moles of sodium bromide (NaBr) present in the solution:

Mass = 20.0 g (given)

Molar mass of NaBr = atomic mass of Na + atomic mass of Br
= (22.99 g/mol) + (79.90 g/mol)
= 102.89 g/mol

Number of moles of NaBr = Mass / Molar mass
= 20.0 g / 102.89 g/mol

Next, we can calculate the number of moles of bromine (Br₂) and sodium chloride (NaCl) produced by the reaction:

Number of moles of Br₂ = Mass of bromine / molar mass of bromine
= 16.0 g / (2 x atomic mass of Br)
= 16.0 g / (2 x 79.90 g/mol)

Number of moles of NaCl = Mass of sodium chloride / molar mass of sodium chloride
= 11.7 g / (atomic mass of Na + atomic mass of Cl)
= 11.7 g / (22.99 g/mol + 35.45 g/mol)

From the balanced equation, we know that for each mole of bromine produced, one mole of chlorine gas is consumed. Therefore, the number of moles of chlorine gas that reacted is equal to the number of moles of bromine produced.

Finally, we can calculate the mass of chlorine gas:

Mass of chlorine gas = Number of moles of Cl₂ x molar mass of Cl₂
= Number of moles of Br₂ x molar mass of Cl₂
= (16.0 g / (2 x 79.90 g/mol)) x (2 x atomic mass of Cl)
= 16.0 g x (atomic mass of Cl / 79.90 g/mol)

By plugging in the values, we can find out the mass of chlorine gas that reacted.

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