For the reaction below, Kp = 1.59 at 100°C. If 1.0 g of SrCO3 is placed in an empty 5.00 L reactor and allowed to reach equilibrium, what will be the pressure in the reactor?

Explanation:

Hello, in this case, one could consider the undergoing chemical reaction as:

Thus, since 1.0 g of strontium carbonate is placed, the equilibrium equation takes the following form, excluding the solid-stated species and considering just the carbon dioxide as it is gaseous:

Hence, since at the beginning there is no carbon dioxide, its pressure at equilibrium equals Kp:

Which was clearly defined above.

Best regards.

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

the answer is 1 i believe