The synthesis of methanol from carbon monoxide and hydrogen gas is described by the following chemical equation:
co(g)+2h2(g)⇌ch3oh(g)
the equilibrium constant for this reaction at 25 ∘c is kc=2.3×104. in this tutorial, you will use the equilibrium-constant expression to find the concentration of methanol at equilibrium, given the concentration of the reactants.

determine the expression for the equilibrium constant, kc, for the reaction by identifying which terms will be in the numerator and denominator:

the equilibrium-constant expression is a mathematical equation that can be rearranged to solve for any of the variables in it. rearrange the equilibrium-constant expression to solve for [ch3oh].
[ch3oh]=[ch3oh]=
1kc[co][h2]2
kc[co][h2]2
kc[co][h2]2
[co][h2]2kc

The expression will be given as:

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as

⇌

The equilibrium-constant expression for the given reaction  is given by:

If we are given with equilibrium constant and equilibrium concentration of carbon monoxide and hydrogen gas we can determine the concentration of methanol at equilibrium.

The expression will be given as:

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

i dont know

explanation: