[The Dude]

How is it that you can predict how many (and what the products will be) if you have more than two reactants? Any sort of info on this matter is appreciated.

[Mitch]

give an example of 2 reactants and we'll try to explain it.

[budullewraagh]

yeah; it's something you have to just be able to do...best way is to compare effective nuclear charges, electronegativities, ionization energies, etc.

[The Dude]

Ok, let's just say you have the following reaction:
NaCO3 + Al2O3 + CH4 ---> CO2 + H2O + ...(then what?)

How do you know this reaction can even occur? What are the products? Since there's 3 reactants, there must be lots of products.

[budullewraagh]

NaCO3 + Al2O3 + CH4 ---> CO2 + H2O + ...(then what?)

well...NaCO3- would definately react with just about anything that can oxidize
i think you were referring to Na2CO3.  personally, i have no idea what the methane would be for...methane is good for...replacing the H atoms with O or halogens, etc.  i dont think youd get a reaction from this system anyway.

[AWK]

This reaction can proceed at very high temperatures (2000 - 3000 K).
At  temperature above 1000 K you can expect decomposition of Na2CO3 and CH4
Na₂CO₃ = Na₂O + CO₂
and CH4 = C + 2H2
At about 1000 K and above carbon dioxide cannot exist at the presence of carbon or hydrogen
so reactions
CO₂ + C = 2CO
and
CO₂ + H₂ = CO + H₂O
can be expected
Moreover Na₂O can react with Al₂O₃ forming NaAlO₂
Finally, depending on ratio of substances and temperature used you can expect:
below 1000 K (approximately) - no reaction or slight decomposition of methane
about 1500 K - CO, H₂O and NaAlO₂,
at much excess of methane reduction CO by hydrogen
CO + H₂ = C + H₂O
and at temperature of order 2500 K and excess of methane you can expect reduction of NaAlO2 by hydrogen followed by reaction Al and Na₂O with formation Al₃C₄ and Na₄C

[The Dude]

How did you come up with that? and where can i find some info so i can figure out these types of reactions by myself?