Question

FeS(s) + 2HCI (al) —> FeCI2(aq) + H2S (s) A reaction mixture initially contains 0.204mol FeS and 0.660mol HCI. Once the reaction reached completion what amount (in moles) of the excess reactant is left?

Answer 1

Step 1

The reaction involved:

FeS(s) + 2HCI (aq) => FeCI2(aq) + H2S (s) (completed and balanced)

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Step 2

Data provided:

0.204 moles FeS and 0.660 moles HCI

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Step 3

The limiting reactant and the excess reactant:

By stoichiometry,

FeS (s) + 2HCI (aq) => FeCI2(aq) + H2S (s)

1 mole FeS --------- 2 moles HCl

0.204 moles FeS --------- X

X = 0.204 moles FeS x 2 moles HCl/1 mole FeS

X = 0.408 moles HCl

For 0.204 moles of FeS, 0.408 moles of HCl are needed, but there are 0.660 moles of HCl.

The excess reactant is HCl

The amount in excess:

0.660 moles of HCl - 0.408 moles HCl = 0.252 moles

Answer: 0.252 moles of HCl