Amixture of ch4 and h2o is passed over a nickel catalyst at 1000. k. the emerging gas is collected in a 5.00-l flask and is found to contain 8.32 g of co, 2.63 g of h2, 42.3 g of ch4, and 49.2 g of h2o. assuming that equilibrium has been reached, calculate kc and kp for the reaction.

                    
According to that Kc is an equilibrium constant in terms of molar concentrations.
and Kc = [C]^c *[D]^d / [A]^a * [B]^b >>>> (1)
in the general reaction:
aA + bB ↔ cC + dD 
and, from our balanced equation:
CH4 + H2O ⇔ Co + 3H2 >>> (2)
So, we need to calculate the concentrations (molarity) of the products and reactants:
the Molarity of CH4 = no. of moles/volume (L)
 and no. of moles = weigh / Molecular weight = 42.3 / 16 = 2.643 moles
so the molarity of CH4 = 2.643 / 5 = 0.528 molar
the molarity of H20 = (49.2 / 18) / 5 =  0.546 molar
the molarity of CO = (8.32/28) / 5 = 0.059 molar
the molarity of H2 = (2.63 / 2) / 5 = 0.263 molar 
By substitution in (1) according to (2);
∴ Kc = [0.059]*[0.263]^3 / ( [0.528]*[0.546]) = 3.7 * 10 ^-3  >>>> (3)
Kp = Kc (RT)^(Δn) >>> (4)
where R is the gas constant = 0.0821,
and Δn is the change in moles in gas= (3(H2) + 1 (CO) - (1 H2O + 1 CH4) = 2
by substition in (4):
∴ Kp = 3.7*10^-3 (0.0821* 1000)^2= 24.939

Only one reactant yields 2 or more products. meaning you will only have 1 product and it will break down into 2 or more products. eg. h2o -> h2 + o2