Fe(s) + S(s) → FeS(s) In one experiment, 7.62 g of Fe are allowed to react with 8.67 g of S.
What is the limiting reagent, and what is the reactant in excess?
Calculate the mass of FeS formed.

The mass of FeS formed : 11.97 g

Reaction

Fe(s) + S(s) → FeS(s)

mass Fe=7.62 g

mol Fe(MW=56 g/mol) :

mass S=8.67 g

mol S(MW=32 g/mol) :

Limiting reactant : smaller ratio(mol:coefficient) ⇒coefficient=1

Fe : S =0.136 : 0.271⇒Fe limiting(smaller), S excess

Mol FeS based on Fe , so mol FeS=0.136

Mass FeS(MW=88 g/mol) :

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-