# Answer to Question #28553 in Organic Chemistry for Mubashir

Question #28553

In one experiment 76.2 gm of Fe are allowed to react with 86.7gm of S. which of the two is limiting reactant.. Fe(x) + S(p) FeS(p) ( Atomic Mass of S =32 a.m.u., Fe= 56 a.m.u.)

Expert's answer

To calculate this, we need to calculate the amount of material of bothreagents

n(Fe) = m / Ar

n(Fe) = 76.2 gm/56 a.m.u. = 1.36 mol

n(S) = m / Ar

n(S) = 86.7gm/32 a.m.u. = 2.7 mol

Starting from the equation:

Fe(x) + S (p) ==> FeS (p)

Starting from the equation, it is seen that the sulfur in excess, thuslimiting reagent is iron.

n(Fe) = m / Ar

n(Fe) = 76.2 gm/56 a.m.u. = 1.36 mol

n(S) = m / Ar

n(S) = 86.7gm/32 a.m.u. = 2.7 mol

Starting from the equation:

Fe(x) + S (p) ==> FeS (p)

Starting from the equation, it is seen that the sulfur in excess, thuslimiting reagent is iron.

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