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.

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.

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.

edit.

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

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?

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.

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?

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

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

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.

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?