Question #c186f

1 Answer
Mar 16, 2016

Here's what I got.


Your starting point here will be the balanced chemical equation for this neutralization reaction.

Hydrobromic acid, #"HBr"#, will react with barium hydroxide, #"Ba"("OH")_2#, to form barium bromide, #"BaBr"_2#, and water

#color(red)(2)"HBr"_text((aq]) + "Ba"("OH")_text(2(aq]) -> "BaBr"_text(2(aq]) + 2"H"_2"O"_text((l])#

As you can see, the reaction consumes #color(red)(2)# moles of hydrobromic acid for every mole of barium hydroxide that takes part in the reaction.

Use the molar masses of the two reactants to determine how many moles of each you're adding to the solution

#16.043 color(red)(cancel(color(black)("g"))) * "1 mole HBr"/(80.912color(red)(cancel(color(black)("g")))) = "0.198277 moles HBr"#

#13.902color(red)(cancel(color(black)("g"))) * ("1 mole Ba"("OH")_2)/(171.342color(red)(cancel(color(black)("g")))) = "0.0811360 moles Ba"("OH")_2#

Now, do you have enough barium hydroxide to ensure that all the moles of hydrobromic acid take part in the reaction?

Use the #color(red)(2):1# mole ratio that exists between the two reactants to find out.

#0.198277color(red)(cancel(color(black)("moles HBr"))) * ("1 mole Ba"("OH")_2)/(color(red)(2)color(red)(cancel(color(black)("moles HBr")))) = "0.0991385 moles Ba"("OH")_2#

As you can see, you don't have enough barium hydroxide to ensure that all the hydrobromic acid takes part in the reaction. This means that barium hydroxide will act as a limiting reagent, i.e. it will determine how much hydrobromic acid actually reacts.

More specifically, this much barium hydroxide will only consume

#0.0811360color(red)(cancel(color(black)("moles Ba"("OH")_2))) * (color(red)(2)color(white)(a)"moles HBr")/(1color(red)(cancel(color(black)("mole Ba"("OH")_2)))) = "0.162272 moles HBr"#

Next, use the #color(red)(2):2# mole ratio that exists between hydrobromic acid and water to determine the theoretical yield of the reaction, i.e. how much water would be produced if the reaction had a #100%# yield.

Since every mole of hydrobromic acid that takes part in the reaction produces one mole of water, you will have #0.162272# moles of water as the reaction's theoretical yield.

Now, use Avogadro's number to convert the given number of molecules of water to moles of water

#7.85 * 10^(25)color(red)(cancel(color(black)("molec. H"_2"O"))) * overbrace(("1 mole H"_2"O")/(6.022 * 10^(23)color(red)(cancel(color(black)("molec. H"_2"O")))))^(color(purple)("Avogadro's number")) = 1.3036 * 10^2"moles H"_2"O"#

As you can see, this value doesn't make much sense in the context of the problem.

The actual yield of the reaction, which is what you actually get after the reaction is completed, must be smaller than the theoretical yield, i.e. the percent yield of a reaction cannot exceed #100%#.

In this case, the reaction can theoretically produce #0.162272# moles of water, but it ends up producing #130.36# moles instead #-># NOT possible.

My guess is that the number of moles of water is incorrect. An alternative would be #7.85 * 10^(22)# molecules. This will be equivalent to #0.13036# moles of water produced by the reaction.

In this case, the percent yield of the reaction, which is defined as

#color(blue)(|bar(ul(color(white)(a/a)"% yield" = "what you actually get"/"what you should theoretically get" xx 100color(white)(a/a)|)))#

will be equal to

#"% yield" = (0.13036 color(red)(cancel(color(black)("moles"))))/(0.162272 color(red)(cancel(color(black)("moles")))) xx 100 = 80.3% -># makes sense