Achemist wants to find kc for the following reaction at 751 k: 2nh3(g) + 3 i2 (g) latex: \longleftrightarrow ⟺ 6hi(g) + n2(g) kc = use the following data at 751 k to find the unknown kc: (1) n2(g) + 3 h2(g) latex: \longleftrightarrow ⟺ 2 nh3(g) kc1 = 0.282 (2) h2(g) + i2(g) latex: \longleftrightarrow ⟺ 2 hi(g) kc2 = 41 enter to 0 decimal places.

The equilibrium constant for the total reaction is

Explanation:

We are given:

We are given two intermediate equations:

Equation 1:

The expression of for the above equation is:

       .......(1)

Equation 2:

The expression of for the above equation is:

      ......(2)

Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.

Now, dividing expression 1 by expression 2, we get:

The above expression is the expression for equilibrium constant of the total equation, which is:

Hence, the equilibrium constant for the total reaction is

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

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