[Barnes123]

Please help me im a totally lost chem student i need help with this question.

Barium sulfate, BaSO4 is made by the following reaction.

Ba(NO3)2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaNO3(aq)
An experiment was begun with 24.90 g of Ba(NO3)2 and an excess of Na2SO4. After collecting and drying the product, 17.1 g of BaSO4 was obtained. Calculate the theoretical yield and percentage yield of BaSO4.

. g of BaSO4 would be expected, hence we have.... % yield

How do i do it. I have no clue please.

[Arkcon]

So you have to determine grams of BaSO₄, given grams of Ba(NO₃)₂.  Ok, how do they differ from one another?  You've been given an equation that relates the two, tell me what are it's units?  You don't see the units symbol, but it's there, understood. Can you guess it?  You realize you'll have to convert, can you do that?

Those are some things to think about.  That's how we help on this forum.  We ask questions to try to laed you to answers on your own.  You'll be better off come exam time, if you crack it yourself.

[pfnm]

You need to compare stochiometric ratios with experimental ratios.

The stochiometric ratios are those listed. Think of them as being the ratio of reagents to each other and reagents to products in a 'perfect' scenario. So in a perfect situation, since you have no coefficients listed for your reagents, this means they are present, in the perfect, or 'stochiometric' scenario (the reaction that goes ahead exactly according to the equation), as one mole Barium Nitrate reacting with one mole Sodium sulfate, to produce one mole barium sulfate and two mole sodium nitrate.

But that's the perfect scenario. For the experiment to go ahead perfectly, you'll need 1:1 ratios of each reactant. It doesn't matter if you have 1 mole of Ba(NO3)2 and 1 mole of Na2SO4, as long as you have the same RATIO.

So you could have half a mole of Ba(NO3)2, and half a mole of Na2SO4 and the reaction would still go to perfect completion.

BUT if you have less of one, or less of the other - then the reaction cannot go to completion. It can only use up the reactant which is present in smallest amounts, and then stop. This reactant, that is present in smallest amounts, is called the 'limiting reactant' or 'limiting reagent'.

How much product you get is determined by your limiting reactant and the ratios between your reactants and your products.

So again, go back to your equation:

Ba(NO3)2(aq) + Na2SO4(aq) ->  BaSO4(s) + 2NaNO3(aq)

Remember that the reactants have a 1:1 ratio. There needs to be as much of one as the other for the reaction to go 'perfectly'. Now look at your products. BaSO4 has the same ratio as the reactants, but NaNO3 has that '2' in front of it. Which means, in the perfect scenario, 2 moles of NaNO3 would be produced from 1 mole of Ba(NO3)2 and 1 mole of Na2SO4.

Remember when I talk of moles in this sense I just mean relationships between the products and the reactants - relative amounts, comparative amounts.

Now, we have been told that sodium sulfate is in excess, that means there is 'more' available. So, the limiting reagent must be barium nitrate.

You need to compare the 'stochiometric' ratio (that is, the ratio of moles that is necessary for a reaction to go to completion, with the experimental ratio (that is, the ratio of moles you actually have).

So you know that once all the Ba(NO3)2 is used up, the reaction ends. Barium nitrate is limiting.

How many moles are there? you have 24.90 grams of barium nitrate. you need to convert this to moles.

Once you know how many moles you have, you know how many moles of product should be obtained: if you got, say, 4.5 moles of barium nitrate (NOTE this is NOT the answer I am using it as an example), you will get the same number of moles of BaSO4. But you'll get twice as many moles of NaNO3, which is clear from your ratios in your equation.

So you know how many moles of barium nitrate you have. This will be the amount of moles of barium sulfate you SHOULD get. Convert the moles you should get, to grams. This is called the theoretical yield.

Compare the amount of grams that were actually formed. This is called the actual yield (17.1 g).

Percent yield = actual yield/theoretical yield x 100

Hopefully this should put you on the right track. If you have any problems see if you can revise stochiometry - these ratios are very important, in fact, getting a basic grasp of stochiometry is very important.

[c0d3]

I got 22.24g as theoretical yield and 76.89% as percentage yield. You may compare  with your answer and see is it the same. Good luck

P.S. Remind me if i am wrong

[pfnm]

24.24g is correct, but percent yield = experimental yield divided by theoretical yield x 100, so 17.1g (as obtained according to the question) divided by 24.24g, gives 0.7054, times 100 gives 70.54 % yield.

[c0d3]

24.24g is correct, but percent yield = experimental yield divided by theoretical yield x 100, so 17.1g (as obtained according to the question) divided by 24.24g, gives 0.7054, times 100 gives 70.54 % yield.

Excuse me but do you get your theoretical yield as 22.24 or 24.24g? Because 17.1/22.24 is 0.7689 and 17.1/24.24 is  0.7054.

[pfnm]

My apologies, I was calculating percent yield with 24.24g as the theoretical yield, rather than 22.24g, which is correct, and hence as you said, 76.89% is right.

[c0d3]

My apologies, I was calculating percent yield with 24.24g as the theoretical yield, rather than 22.24g, which is correct, and hence as you said, 76.89% is right.

No need to apologize, this is a place for us to learn. o matter we are 18 years old teen or 80 years old golden citizen. we still have to learn. Cheers and YaY