lron(Ill) sulfide reacts with carbon monoxide according to the equation: Fe2O3(s

Question

lron(Ill) sulfide reacts with carbon monoxide according to the equation:
Fe2O3(s) + 3 CO(g)--> 2 Fe(s) + 3 CO2(g)
A reaction mixture initially contains 22.55 g Fe2O3 and 14.78 g CO. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left? lron(Ill) sulfide reacts with carbon monoxide according to the equation:
Fe2O3(s) + 3 CO(g)--> 2 Fe(s) + 3 CO2(g)
A reaction mixture initially contains 22.55 g Fe2O3 and 14.78 g CO. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?
Fe2O3(s) + 3 CO(g)--> 2 Fe(s) + 3 CO2(g)
A reaction mixture initially contains 22.55 g Fe2O3 and 14.78 g CO. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?
A reaction mixture initially contains 22.55 g Fe2O3 and 14.78 g CO. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

Explanation / Answer

Fe2O3(s)+3CO(g) -->2Fe(s) +3CO2(g)
if a particular reactant is left even after the reaction that definitely means the other reactant that was completely used up was limited and the reaction has taken place according to the no. of moles of this limiting reactant.

so first of find the limiting reactant!
no. of moles of Fe2O3 = 22.55/ ( 112 +48) =22.55 / 160 =0.141mol
no. of moles of CO = 14.78/ (12+16) =14.78/28 =0.527mol

now look at the equation ,the mole ratio of Fe2O3 ; CO = 1: 3
now since we have 0.527mol of CO , the no. of moles of Fe2O3(s) REQUIRED according to the stoichiometry is (0.527/3) that is 0.1756mol but theres only 0.141mol of Fe2O3
so from that it is claer that Fe2O3 is limited.

therefore since Fe2O3 is the limiting reactant ,it will be completely used up and CO will be left;
no. of moles of Fe2O3 reacted = 0.141mol
according to stoichiometry no. of moles of Co reacted = 0.141*3 =0.423 mol
therefore no. of moles of CO remaining = 0.527 - 0.423mol = 0.104mol
mass of CO left = 0.104 * 28 =2.912g

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