[ssssss]

CaC2 + 2H2O---> C2H2 + Ca[OH]2

C2H2 + H2---->C2H4

nC2H4---->[C2H4]n

If we got 10 Kg of CaC2 then How much of [C2H4]n will be formed.

Now what i did was first combined the 3 equations.

nCaC2 + 2nH2O + nH2--->[C2H2]n + nCa[OH]2

Moles of CaC2=10x1000/64

Now what to do next?

[Mitch]

Having that "n" there makes the problem look more scary. Don't worry about trying to use it in your equations. The "n" acts as a multiplication term when it is in front of the compund or whether it is a subscript of the compound.  The "n" in the third equilibrium cancel each other out since they both multiply the compound the same amount, so ignore it and treat the problem the way you normally would.

This is an advanced way of looking at the problem and is not recommended. Maybe someone else can give you a better approach.  

[ssssss]

Heres a similar question



2CoF2 + F2------>2CoF3 and followed by,
[CH2]n + 4nCoF3------>[CF2]n + 2nHF + 4nCoF2

1 Kg of [CF]n are formed.How much Flourine will be cinsumed?

[Mitch]

Again the n can be factored out, so don't worry about it.