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Iron(II) sulfide reacts with hydrochloric acid according to the reaction:

FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)

A reaction mixture initially contains 0.240 mol FeS and 0.646 mol HCl.

what amount (in moles) of the excess reactant is left?

Verified Answer :

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The balanced chemical reaction would be:

FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g) 

We are given the amount of the reactants to be used for the reaction. We use these amounts. First, we determine the limiting reactant of the reaction. From the data, we can say that FeS is the limiting ad HCl is the excess reactant. We calculate as follows:

Amount of HCl used: 0.240 mol FeS x 2 mol HCl / 1 mol FeS = 0.48 mol HCl

0.646 - 0.48 = 0.166 mol HCl left

Answer: The amount (in moles) of the excess reactant left is, 0.166 mol

Explanation : Given,

Moles of FeS = 0.240 m ol

Moles of HCl = 0.646 mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

\(FeS(s)+2HCl(aq)\rightarrow FeCl_2(s)+H_2S(g)\)

From the balanced reaction we conclude that

As, 1 mole of \(FeS\) react with 2 mole of \(HCl\)

So, 0.240 moles of \(FeS\) react with \(0.240\times 2=0.48\) moles of \(HCl\)

From this we conclude that, \(HCl\) is an excess reagent because the given moles are greater than the required moles and \(FeS\) is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of \(HCl\)

Remaining moles of HCl = 0.646 - 0.48 = 0.166 mol

Thus, the amount (in moles) of the excess reactant left is, 0.166 mol

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