I noticed Michael posted a solution while I was writing mine. I will post anyways and think about his answer after.

**Puzzle 68/69** Let's start by defining an order on the set of complexes \\((a,b,c) \text{ where } a,b,c \in \mathbb{N}\\) with a representing \\(\text{H}\\), b representing \\(\text{O}\\) and c representing \\(\text{H}_2\text{O}\\) without taking the reaction into account. We say that:

$$ (a,b,c) \le (a',b',c') \text{ if } a \le a' \text{ and } b \le b' \text{ and } c \le c' $$

We can interpret this as meaning: Given a complex \\(C' = (a',b',c')\\), I can get the sub-complex \\(C = (a,b,c)\\) if all compontents of \\(C\\) are present in \\(C'\\). This ordering forms a symmetric monoidal poset under addition of natural numbers with the unit \\((0,0,0)\\).

Now we can add the reaction by demanding, that anytime we have a multiple of \\(2\text{ H} + \text{O}\\) we can get that many \\(\text{H}_2\text{O}\\). Thus we add additional ordering given by:
$$ (a-2n,b-n,c+n) \le (a,b,c) \text{ for } n \in \mathbb{N} \text{ and } a \ge 2, b \ge 1$$

Given this additional structure, we still have at least a preorder since:

$$ (a-2n,b-n,c+n) \le (a-2n,b-n,c+n) \text{ (reflexivity) } $$
$$ (a-2n,b-n,c+n) \le (a,b,c) \text{ and } (a,b,c) \le (a',b',c') \text{ implies } (a-2n,b-n,c+n) \le (a',b',c') \text{ (transitivity) } $$

We even retain the poset structure since given \\((a-2n,b-n,c+n) \le (a,b,c)\\) we can't also have \\((a,b,c) \le (a-2n,b-n,c+n)\\) for \\(n \ge 1\\) since \\(a \not\le a-2n\\) and \\(b \not\le b-n\\) and in the case \\(n=0\\), \\((a,b,c) = (a,b,c)\\).

*Edit: corrected mistaken \\(c \not\le c+n\\) and \\(n \le 1 \\)in the above section*

It remains to be shown that this new poset also forms a symmetric monoidal poset under addition of natural numbers with the unit \\((0,0,0)\\).

$$ \text{Unit laws hold since: } (0+a-2n,0+b-n,0+c+n) \le (a-2n,b-n,c+n) \le (a-2n+0,b-n+0,c+n+0)$$
$$ \text{Associativity holds since: } ([(a-2n)+(a'-2n')]+(a''-2n''),[(b-n)+(b'-n')]+(b''-n''),[(c+n)+(c'+n')]+(c''+n'')) $$
$$ \le ((a-2n)+[(a'-2n')+(a''-2n'')],(b-n)+[(b'-n')+(b''-n'')],(c+n)+[(c'+n')+(c''+n'')]) \text{ for all } a,a',a'',b,b',b'',c,c',c''$$

This leaves: \\( x \le x' \textrm{ and } y \le y' \textrm{ imply } x \otimes y \le x' \otimes y' \\). Our new structure provides two extra cases:
$$ (a-2n,b-n,c+n) \le (a,b,c) \text{ and } (d-2m,e-m,f+m) \le (d,e,f) \text{ should imply } ((a-2n)+(d-2m),(b-n)+(e-m),(c+n)+(f+m)) \le (a+d,b+e,d+f)$$
This is the case since:
$$((a-2n)+(d-2m),(b-n)+(e-m),(c+n)+(f+m)) = ((a+d)-2(n+m),(b+e)+(n-m),(c+f)+(n+m)) \le (a+d,b+e,d+f) $$
By taking \\(a'=a+d, b'=b+e, d'=d+f, n'=n+m\\)

and

$$ (a-2n,b-n,c+n) \le (a,b,c) \text{ and } (d,e,f) \le (d',e',f') \text{ should imply } (a-2n+d,b-n+e,c+n+f) \le (a+d',b+e',c+f')$$
We can do this component-wise:
$$a \le a \text{ and } d \le d' \text{ in our original symmetric monoidal poset implies: } a+d \le a+d' \text{ so: } a-2n+d \le (a+d')$$
$$b \le b \text{ and } e \le e' \text{ in our original symmetric monoidal poset implies: } b+e \le b+e' \text{ so: } b-n+e \le (b+e')$$
$$c+n+f \le c+f' \text{ since } c+n \le c \text { for all } n \text { and we know } f \le f'$$
(This last line seems somewhat circular to me, so I'm not sure if it's correct...)

Finally, symmetry is guaranteed by the fact that both addition an subtraction of natural numbers are commutative. Thus, provided the above arguments are correct we indeed have a symmetric monoidal poset.

*Edit: changed preorder to poset*