If 25.0g solid iron (III) sulfide reacts with 25.0g gaseous hydrogen chloride to form solid iron (III) chloride and 5.61g hydrogen sulfide gas, what is the percent yield of the reaction?

(Please show work)

Explanation:

Hello!

In this case, since the described chemical reaction is:

We can see the 6:2 mole ratio between hydrogen chloride (molar mass = 36.46 g/mol) and hydrogen sulfide (molar mass = 34.09 g/mol) and 1:3 between iron (III) sulfide (molar mass = 207.91 g/mol) and hydrogen sulfide. In such a way, we can compute the yielded moles of hydrogen sulfide by each reactant in order to identify the limiting one:

It means that the HCl is the liming reactant as it produces less moles of hydrogen chloride than the iron (III) sulfide. Therefore, the theoretical yield in grams is:

Thus, the percent yield is:

Best regards!

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.