Wasteless Potential

No of reagents required:13
No of resulting products:2

The process is split into 8 parts for 5 arms (4 piston arms and 1 regular arm)
Note:the simple arm (aka arm 5) has only one role:to get 3 of the 4 QS required (2 are given by arm 3 and 1 by decomposing one of the reagents).(the 4th QS is brought by arm 4)

Part 1:requires 2 reagents
Arm 1:grabs one reagent, decomposes it into 3 QS and 1 lead.Leaves all of the QS for arm 3. Deposits the lead at the glyph of purification. Returns to the initial position
Arm 2:grabs another reagent, decomposes it into 2 QS and 1 lead-QS triple bonded element. With the help of 4 QS (1 from arm 4 and 3 from arm 5, from which 2 are given by arm 3), we can turn the lead into silver. Deposits the Silver-QS triple bonded element to a glyph of bonding for the next part.Returns to the initial position
Arm 3:gives 2 of the 3 QS to arm 5, deposits the 3rd QS for later parts
Arm 4:uses QS to transform lead into tin
Arm 5:uses 3 QS to transform tin into silver

Part 2:requires 2 reagents
Arm 1:grabs reagent, decomposes it into 3 QS and 1 lead. Leaves all QS for arm 3. Deposits the lead at the glyph of purification (this lead and the lead from 1st part make into tin). Deposits tin somewhere nearby. Returns to initial position
Arm 2:does the exact same thing as in part 1 with a singular change.With the help of arm 4, bonds the first 2 Silver-QS triple bonded elements together with a simple bond.Returns to the initial position afterwards
Arm 3:does the exact same thing as in part 1
Arm 4:does the exact same thing as in part 1,then helps arm 1 bonding the 2 Silver-QS triple bonded elements,then repositions the result for the next bond, then returns to the initial position
Arm 5:does the exact same thing as in part 1

Part 3:requires 2 reagents
Arm 1:does the same thing as in part 1
Arm 2:does the same thing as in part 2
Arm 3:does the same thing s in the previous parts
Arm 4:does the same thing as in part 2, but... repositions the new result for the bond with a QS that happens in the next part
Arm 5:(i am no longer going to repeat what arm 5 is doing)

Part 4:requires 1 reagent
Arm 1:grabs reagent, decomposes it into 3 QS and 1 lead. 2 of the QS will be used by arm 1 and the third by arm 2.Deposits lead at glyph of purification (2 lead=tin). Deposits the 2 atoms of tin into the glyph of purification to make iron. Then with iron and 2 QS, arm 1 makes silver, which is now going to be bonded with the rest of the product. Returns to th initial positian afterwards
Arm 2:steals 1 QS, bonds QS with the repositioned element. Brings new element in for the final bond with the silver made by arm1, then deposits product into the product area, then returns to the initial position.
Arm 3:*proceeds to imitate Dio's throwing knife scene from part 3*
Arm 4:After repositioning the element from part 3,waits for arm 2 to bond the QS and take the resulting element for the final bond, then returns to the initial position
Arm 5:same as arm 3

Part 5:requires 2 reagents
Arm 1:repeat of part 1 except: gives only 2 QS to arm 3, uses lead and QS to make tin instead of depositing it to the glyph of purification
Arm 2:repeat of part 1
Arm 3:gives the 2 QS to arm 5 before returning to the initial position
Arm 4:same as part 1

Part 6:requires 2 reagents
Arm 1:repeat of part 5 except:ead to glyph of purification, uses QS to make iron
Arm 2:repeat of part 2
Arm 3:repeat of part 5
Arm 4:repeat of part 2

Part 7:requires 2 reagents
Arm 1:repeat of part 6 except:2 lead=1 tin, iron+QS=copper
Arm 2:repeat of part 3
Arm 3:repeat of part 5
Arm 4:repeat of part 3

Part 8:requires the extra 3 QS deposited by arm 3 throughout the run
Arm 1:swaps copper and tin's places. uses 2 QS that were deposited by arm 3 to make tin into copper. uses the 2 copper to make silver. rest is same as part 4
Arm 2:repeat of part 4 (the only difference is tht the QS was one of the 3 QS deposited by arm 3 throughout the run)
Arm 3:brings the 3 QS to be used by arm 1 and arm 2
Arm 4:repeat of part 4
Arm 5:repeat of part 4

TLDR:13 reagents=13 lead+39 QS. (1 lead+4 QS)*3+4 QS+(4 lead+2 QS)=(1 lead+4 QS)*3+4 QS+(3 lead+5 QS)=product
PS:i know this needs some optimizations tbh, but that's the best i could do, and i prise the ones who did better