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H.5.b

Posted: Thu Apr 05, 2018 8:59 pm
by Alexander Hari 1L
I am having trouble figuring out how to balance part H.5.b. Which is: Mg(N3)2(s)+H2O(l)-->Mg(OH)2(aq)+HN3(aq)..
I am not sure how we are suppose to distribute the numbers following the parenthesis. Do we multiply it out so? For example, would Mg(N3)2 become MgN6? Or is there some other way we are suppose to distribute this? Thanks!

Re: H.5.b

Posted: Thu Apr 05, 2018 9:48 pm
by Nicole Shak 1L
Yes, Mg(N3)2 would become MgN6. This just means that there are 6 N, so you would also need 6 N on the other side of the equation. When there's a number outside the parentheses you multiply that number by the subscript in the parentheses (so if you had (H6)2 in subscripts it would be the same as H12). Hope that makes sense!

Re: H.5.b

Posted: Fri Apr 06, 2018 11:15 am
by Elizabeth Parker 1K
Yea just think of it as being a simple algebraic problem. You distribute and then you can continue the problem.

Re: H.5.b

Posted: Fri Apr 06, 2018 12:02 pm
by Alexis Bravo 1D
Yes you would just multiply the the N3 by 2 and the OH by 2 in order to be able to continue balancing the equation

Re: H.5.b

Posted: Fri Apr 06, 2018 12:26 pm
by Jesus A Cuevas - 1E
As a sidenote, for those that read like 4(O2)3, you would first multiply the subscripts then multiply by the coefficient. In this example, you'd get 6 oxygen from multiplying the subscript (2 x 3= 6), then multiply the result by the coefficient (4 x 6 = 24) giving you a total of 24 Oxygens.

Re: H.5.b

Posted: Fri Apr 06, 2018 5:25 pm
by Haison Nguyen 1I
When there's a subscript following the parenthesis, you multiply that number by the number in the parenthesis. So yes Mg(N3)2 would be Mg(N6) and you ignore the MG since the subscript doesn't apply to that element. So since you have 6N on the left side, you try to get 6N on the right side as well.

Re: H.5.b  [ENDORSED]

Posted: Sat Apr 07, 2018 1:54 pm
by Chem_Mod
Mg(N3)2 is written in such a way because (N3)- is a polyatomic ion and there needs to be 2 units of these for this Mg compound. It is different from MgN6 as earlier posts seem to suggest.

For purpose of balancing equations, Mg(N3)2 does contain 1 Mg and 6 N atoms.

Re: H.5.b

Posted: Tue Apr 10, 2018 3:17 pm
by 204929947
Yes, you would have to multiply the subscript by the parenthesis. The atoms in the parenthesis show you that you would only have to multiply the subscript by what is inside the parenthesis. It's like in math when you distribute a coefficient just that here, the coefficient is at the end of the parenthesis.

For example, if you have PtCl2(NH3)2
You can see that Platinum and Chlorine are not in the parenthesis, therefore there are only one chlorine and one platinum
Nitrogen and Hydrogen are both inside the parenthesis, followed by the number 2
So you would multiply one nitrogen atom by 2 (the subscript) which will give you 2 nitrogen atoms and then, since you have three hydrogens (H3) and the subscript is two, you multiply 2x3 which will give you 6 atoms of hydrogen.
so all together you have
1 platinum, 1 chlorine, 2 nitrogen, and 6 hydrogen atoms