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?

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:

From the balanced reaction we conclude that

As, 1 mole of react with 2 mole of

So, 0.240 moles of react with moles of

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

Now we have to calculate the excess moles of

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

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.

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