Question

According to the reaction $$Pb(s) + S(s) \to PbS(s)$$ , when 20.7 grams of lead are reacted with 6.4 grams of sulphur:
A.There will be excess of 20.7 grams 04 lead.
B.The sulphur will be in excess by 3.2 grams.
C.The lead and sulphur react completely without any excess reactants.
D.The sulphur will be the limiting factor in the reaction.
E.There will be an excess of 10.35 grams of lead.

Answer 1

Hint: Limiting reagent is defined as the substance which has been fully absorbed during the chemical reaction. The reaction tends to come at the end when the limiting reactant is completely consumed. A fixed amount of the reactants is to be needed for the completion of reaction according to the concept of stoichiometry. They are also known as limiting reactants or the limiting agent.

Complete step-by-step answer:So now we must know the atomic mass of each of the element which is used in the reaction to be carry out that is Pb and S. so the atomic mass of Pb is 207 $$gmo{l^{ - 1}}$$and S is 32$$gmo{l^{ - 1}}$$. So now let us find out the moles of each element used. So the formula which we will be using for finding moles is the following:
Number of Moles= $$\dfrac{{mass given}}{{molar mass}}$$
So let us find out the moles for Pb first. The given mass of Pb is 20.7 and molar mass is 207. So on substituting the values we get,
Number of moles=$$\dfrac{{20.7}}{{207}} = 0.1$$
Now let us find the number of moles os S .
Number of moles=$$\dfrac{{6.4}}{{32}}$$=0.2
So we see that the moles of sulphur is more than that of lead so sulphur is an excess reactant while Pb is the limiting reactant. So out of 0.2 moles of sulphur only 0.1 mole of it will react with Pb. So 0.1 moles of sulphur would be left. So now let us calculate the remaining moles in grams :
Mass of excess S= $$0.1 \times 32$$=3.2g
So the sulphur will be excess by 3.2g.

So the correct answer is option B.

Note:The limiting reagent is generally calculated for knowing the percentage yield of the reaction. The limiting reactant determines when the reaction would stop. We calculate the exact amount of the reactant that is to be used by the reaction stoichiometry. The limiting reagent depends on the mole ratio rather than the masses of the reactant that has been used.