Question

How much magnesium sulphide can be obtained from 2.00g of Magnesium and 2.00g of Sulphur by the reaction $Mg + S \to MgS$? Which is the limiting agent? Calculate the amount of one of the reactants which remains unreacted. $IMg = 24$, $S = 32J$

Answer 1

Hint: The amount of substance present in a given sample is the mole of a substance. To calculate the number of moles, the following equation can be used
$no\;{{of moles = }}\dfrac{{given\;{{mass}}}}{{molar\;{{ mass}}}}$
When 2 substances react under specific conditions, products are formed.
The substance that has a lesser number of moles can ask as the limiting agent. Since it has a lesser number of moles, it will be completely consumed after the chemical reaction completes.

Complete step by step answer:
Here 2 g of Magnesium (Mg) and 2 g of Sulphur (S) are reacting to give the product magnesium sulphide. Let us now find out the number of moles of each substance reacting
No of moles of Magnesium $ = \dfrac{2}{{24}} = \dfrac{1}{{12}} = 0.0833$ as the molar mass of Magnesium is 24.
We know the molar mass of Sulphur is 32 and therefore, the number of moles of Sulphur reacting is $\dfrac{2}{{32}} = \dfrac{1}{{16}} = 0.0625$
The number of moles of Sulphur is less. So, we can say that Sulphur is the limiting agent and it is completely consumed in the course of the reaction.
Therefore, we can say that $0.0625$ the moles of MgS will be formed.
From the number of moles, the mass of the substance can be deducted.
$no\;{{of moles = }}\dfrac{{given\;{{mass}}}}{{molar\;{{ mass}}}}$,
$mass = molar\;{{ mass }} \times {{no \;of\; moles}}$
$ = 56 \times 0.0625$
$ = 3.5g$
$3.5g$ of MgS is produced.
We know that as Sulphur is the limiting agent, it is completely reacted. But Mg will remain.
So, moles of Magnesium left unreacted is $(0.0833 - 0.0625)moles$
$ = 0.0208moles$
Therefore, the mass of Magnesium remaining $ = molar\;{{ mass }} \times {{no \;of \;moles}}$
$ = (0.0208 \times 24)g$
$ = 0.4992g$
Thus 0.5 g

Note: An alternative method to solve this is we know that 32g of Sulphur reacts with 24g of Magnesium to give 56g of MgS. So we can say that
 24 g magnesium $ \to $32 g of Sulphur.
So, $2g\;{{Mg}} \to \dfrac{{32}}{{24}} \times 2$
2g Mg $ \to $2.6 g of Sulphur
Thus, 2 g of Mg should react with 2.6 g of Sulphur. But there is only 2g of S, so it is the limiting agent.
$32g\;{{Sulphur }} \to {{24 g \;of \;Magnesium}}$
$2{{ g \;of\; Sulphur = }}\dfrac{{24}}{{32}} \times 2$
2g of Sulphur reacts with 1.5 g of Mg.
$2g\,{{Sulphur }} \to {{1}}{{.5g of Mg}}$
Therefore, Mg remaining unreacted is $(2 - 1.5)g = 0.5g$