# stoichiometry

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When Ba(NO3)2 and K2CrO4 (sorry, i don't know how to do subscripts, but all the numbers should be subscripts!) react in aqueous solution, the yellow solid BaCrO4 is formed. Calculate the mass of BaCrO4 that forms when 3.50 x 10^(-3) mol of solid Ba(NO3)2 is dissolved in 265 mL of 0.01000 M K2CrO4 solution.

I DON'T GET IT.
i can't even begin to solve it eek.

• stoichiometry -

Write and balance the equation.
Determine mols Ba(NO3)2 (given in the problem).
Calculate mols K2CrO4 from mols = M x L.
Determine mols product formed.
Convert that to grams.

• stoichiometry -

1. balance the equation (they give you reactants and product(s))

You have to then find the limiting reagent, which would then form the product.(find moles product theoretically formed from each starting material)

3.50x10^-3 mol Ba(NO3)2(__mol BaCrO4/__mol Ba(NO3)2)= ___mol BaCrO4 formed from Ba(NO3)2 theoretically

to find the amount of moles of BaCrO4 formed from the K2CrO4 you need to convert the 265ml to L thne use the Molarity of the solution given (M=mol/L) and convert that to moles with the ratio(moles of K2CrO4 and moles BaCrO4)you found before after
balancing the equation.

__L K2CrO4(0.01000mol/L K2CrO4)(___molBaCrO4/____molK2CrO4)=_____mol BaCrO4 formed from K2CrO4 theoretically

The number of moles of BaCrO4 formed from one of the reactants that is less is the limiting reagent. Take the moles of BaCrO4 formed from the limiting reagent and then convert that to grams.

___mol BaCrO4(1molBaCrO4/____g BaCrO4)= __________g BaCrO4

If you need help or have questions just ask.

• stoichiometry -

edit.
___mol BaCrO4(_________g BaCrO4 /1molBaCrO4)= __________g BaCrO4

• stoichiometry -

i still don't really get it because i have no clue where that equation is coming from. where do you come up with that?

• stoichiometry -

The first step is write and balance the equation.
Ba(NO3)2 + K2CrO4 ==> BaCrO4 + 2KNO3.

I can give you a link if you don't know how to do that.

• stoichiometry -

i think i get that. and then should i reduce it to the net ionic?

Ba(2+) + CrO4(2-) -> BaCrO4

or should i leave it how it is?

• stoichiometry -

Leave it as it is. Now you have mols Ba(NO2)2 from the problem. Calculate mols K2CrO4 from M x L.

• stoichiometry -

i'm confused because i don't get why the mols wouldn't be the same. the ratio of Ba(NO3)2 to K2CrO4 is 1:1, so if there are 3.50 x 10^(-3) mol of Ba(NO3)2, wouldn't there be 3.50 x 10^(-3) mol of KrCrO4?

• stoichiometry -

C'mon. Read the problem. They aren't the same because the problem CHOSE to make them different. Let's look at something different. For example, suppose we want to mix 100 g water with 50 g NaCl. According to your reasoning, I would be forced to use either 50 g water or 100 g NaCl so they would be the same. The reactants in ANY reaction can be, and usually are, different. In the problem you originally posted, we could just as easily had 5.6 x 10^-4 mols Ba(NO3)2 and we added 500 mL of 0.0011 M K2CrO4. Or switch the numbers around. Or make them anything you wish. It is true, as you point out, that the ratio is 1 mol Ba(NO3)2 to 1 mol K2CrO4 to 1 mol BaCrO4 to 2 mols KNO3 BUT who says they must be the same unless we just choose to make them the same. Just because they react in the ratio of 1:1:1:2 doesn't means we MUST put 1 mol Ba(NO3)2 and 1 mol K2CrO4. The will always react in a 1:1 ratio, no matter how much I mix together. And they did. 0.00265 mols Ba(NO3)2 reacted with 0.00265 mols K2CrO4 (1:1) to form 0.00265 mols BaCrO4 and 2*0.00265 mols KNO3. When all the shouting was over, we had what in the beaker? We had 2*0.00265 mols KNO3, we had 0.00265 mols BaCrO4, we had NO K2CrO4 (because all of it reacted)(actually we had a little because BaCrO4 has a VERY SMALL solubility) and we had 0.00350 - 0.00265 = 0.00085 mol Ba(NO3)2.

• stoichiometry--big fix. -

Well another correction...stupid mistake of mine (slap forehead)

Just use this not the above...(just removed something

balance the equation (they give you reactants and product(s))

3.50x10^-3 mol Ba(NO3)2(__mol BaCrO4/__mol Ba(NO3)2)= ___mol BaCrO4 formed from Ba(NO3)2 theoretically

___mol BaCrO4(_________g BaCrO4 /1molBaCrO4)= __________g BaCrO4

• stoichiometry--big fix. -

wouldn't the mols of Ba(NO3)2 completely cancel out? and then the equation would just be _mol BaCrO4 = mol BaCrO4, which confuses me. =/
and same with the second equation.

• stoichiometry--big fix. -

Um..just listen to Dr.Bob (I've confused myself very much now) I'm sorry.

• stoichiometry--big fix. -

haha, okay.
well, thank you for your help!

• stoichiometry--big fix. -

This is what happens when one multitasks, unfortunately. =|

• stoichiometry -

You have ONE reaction between Ba(NO3)2 and K2CrO4. It is.
Ba(NO3)2 + K2CrO4 ==> BaCrO4 + 2KNO3.
I don't know where you get the canceling out thing. mols Ba(NO3)2 = 3.5 x 10^-3 from the problem. Put that under Ba(NO3)2 on your scratch pad. Now how many mols K2CrO4 do you have? You have mols = M x L. M = 0.01000 and L = 0.265 so M x L = 0.0100 x 0.265 = 0.00265 mols. Put that under K2CrO4. Now compare the numbers. You have
0.00350 mols Ba(NO3)2 reacting with 0.00265 mols K2CrO4 and it will produce 0.00265 mols of BaCrO4 (since you correctly pointed out that it was 1 mol Ba(NO3)2 to 1 mol K2CrO4 to produce 1 mol BaCrO4. (What happened to the remainder of Ba(NO3)2?) It is simply unreacted. So you have formed 0.00265 mols BaCrO4. Convert that to grams BaCrO4 from grams = mols x molar mass.

• stoichiometry -

Actually I think that my method was correct.(Just makeing sure though, I am posting how I did it)

So actually focusing on the problem and solving this on paper..

1 Ba(NO3)2 + 1 K2CrO4 => 1 BaCrO4+ 2 KNO3

3.50x10^-3mol Ba(NO3)2

265ml = 0.265L
0.265L (0.0100mol/L K2CrO4)= 0.00265mol K2CrO4

I used those and then solved for the limiting (even though if I think about it, it there is no math involved, but I need to see it-visual learner)

3.50x10^-3 Ba(NO3)2 (1mol BaCrO4/1molBa(NO3)2= 0.00350 BaCrO4

0.00265 K2CrO4(1mol BaCrO4/1mol K2CrO4)= 0.00265 mol BaCrO4 (Limiting)

so using that...
0.00265 mol BaCrO4 (253.3216g BaCrO4/
1mol BaCrO4)= 0.6713g BaCrO4

• stoichiometry -

Christina, You are right that K2CrO4 is the limiting reagent but you need not go through all of the calculations to show that. If you have 0.00350 mol Ba(NO3)2 reacting with 0.00265 mols K2CrO4, (and it's 1:1 as it is here) it should be obvious that it can form only 0.00265 mol BaCrO4 with the difference (0.00350 - 0.00265 = 0.00085 mol) Ba(NO3)2 remaining in solution unreacted.

• stoichiometry -

Yes, maybe I'm slow then.

I see that the leftover unreacted Ba(NO3)2 is 0.00085 mols but I had gotten confused with the KNO3 being another product of the reaction. And since you did not include K2CrO4 in the calculations above I was confused..it took me awhile to figure out that:
It was 1:1 so Ba(NO3)2 and the K2CrO4 would react in the same ammount together and thus there would be some left over of Ba(NO3)2 =D

I have to say that my thinking is slower due to the hyperthyroidism I have right now (concentrating is difficult ;___; but at least I got the problem)