## H.5

April Meza 3K
Posts: 66
Joined: Sat Sep 21, 2019 12:36 am

### H.5

(b) Mg(N3)2(s) + H2O(l) ---> Mg(OH)2(aq) + HN3(aq)

Can someone explain how to balance this equation? I get stuck after a few steps. Thanks!
Last edited by April Meza 3K on Thu Oct 03, 2019 9:50 pm, edited 1 time in total.

ahuang
Posts: 52
Joined: Fri Aug 30, 2019 12:17 am
Been upvoted: 1 time

### Re: H.5

First, you notice that there are 6N on the left and only 3N for products. Thus, multiply HN3 by 2. After, there are 2H for reactants and 4 for products. Therefore, multiply H2O by 2. This will finalize balancing the equation:
Mg(N3)2 + 2H2O -> Mg(OH)2 + 2HN3

Lara Kristine Bacasen 1L
Posts: 110
Joined: Sat Jul 20, 2019 12:16 am
Been upvoted: 1 time

### Re: H.5

If you really have trouble balancing it out I suggest you try writing out what elements are there and counting out what is on each side and have a side by side comparison of the elements on the product and reactant side. That is what I did when starting out. You should first try to look at the combined elements and leave the ones that are individual for later (for example do FeS first instead of H2 or something). If they are all combined into molecules like this problem I suggest leaving O last (something I remember doing for chem in highschool)
Okay so the problem. I read it left to right
We look at the larger atoms and see that Mg is balanced. so no change. Next I look at N and see there are six on the reactant side but three on the product side. This means we need to balance it out by putting a stoichiometric coefficient of two in front of HN3 on the product side to make it 6 N atoms.
Alright so next I look at H atoms and see there is two H's on the reactant side and now there is four on the product side because we placed a 2 in front of HN3. We need to balance it again but this time the stoichiometric coefficient is on the reactant side in front of the H20. We need to place a 2 in front balance it out. Once that is done we now check O atoms and see there are two on each side. Now it is balanced. In the end do a comparison of who many atoms of each element are on each side and now you're done.