Calculate the number of milliliters of 0.570 M Ba(OH)2 required to precipitate all of the Cr2+ ions in 140 mL of 0.790 M Cr(NO3)2 solution as Cr(OH)2. The equation for the reaction is:

Cr(NO3)2(aq) + Ba(OH)2(aq) Cr(OH)2(s) + Ba(NO3)2(aq)

Question

Calculate the number of milliliters of 0.570 M Ba(OH)2 required to precipitate all of the Cr2+ ions in 140 mL of 0.790 M Cr(NO3)2 solution as Cr(OH)2. The equation for the reaction is:

Cr(NO3)2(aq) + Ba(OH)2(aq) Cr(OH)2(s) + Ba(NO3)2(aq)

joannablackwelder · Answer

First, count all reactant element and product elements. Numbers balance with 1 mol of each ionic compound. So, we can now convert. 140 mol Cr(NO3)2 *(1L/1000mL)*(0.790 mol Cr(NO3)2/1L)*(1 mol Ba(OH)2/1 mol Cr(NO3)2)*(1L/0.570 mol Ba(OH)2)*(1000 mL/1L)= 194 mL of Ba(OH)2. Do you see how al the other units cancel out?

anonymous · Answer

Yes. I was just wondering when they asked in the question "140 mL of 0.790 M Cr(NO3)2 solution as Cr(OH)2" what exactly do they mean by "as Cr(OH)2?"

anonymous · Answer

Does that even related to the problem?

joannablackwelder · Answer

I thin what they mean is that the other ions will still be floating around in the aqueous solution while the Cu(OH)2 precipitates out of solution.

anonymous · Answer

But with that being said you don't need to use the mole or molar mass of Cu(OH)2 right? (It's just confusing since they included in the question if it doesn't need to be part of the solution)

joannablackwelder · Answer

Why would you not?

anonymous · Answer

Well, the reason i'm asking is because I don't get it so I don't know why I wouldn't use it.

anonymous · Answer

Someone already made the calculations and they didn't use it

joannablackwelder · Answer

It is true that you don't need the molar mass because you are not converting  to mass, but you do need the molar ratio between the reactants.

joannablackwelder · Answer

*Between the reactants and products

anonymous · Answer

But when you did it, I don't see you using the mole ratio of Cu(OH)2, unless I've missed it

joannablackwelder · Answer

I did use the mol ratio. Since i is 1:1 it desn't affect anything and is easy to miss. Try writing out the conversion I typed.

joannablackwelder · Answer

*Since it

anonymous · Answer

I wrote it out but don't see where you used it. The answer is correct. I'll figure that part out myself. Thank you very much for the help!

joannablackwelder · Answer

I used it when I multiplied by 1 mol Ba(OH)2/1 mol Cr(NO3)2. Hope that helps!