Question

How much Magnesium sulphide can be obtained from 2.00 G of Magnesium and 2.00 g of sulphur by the reaction, Mg + S --> MgS? Which is the limiting agent? Calculate the amount of one of the reactants which remains unreacted.

Answer 1

$\sf{\underline {We\:shall\:convert\:the\:masses\:into\:moles.}}$

$\sf{2.00 \: g \: of \: Mg = \frac{2.00}{24} = 0.0833 \: moles \: of \: Mg}$

$\sf{2.00 \: g \: of \: s = \frac{2.00}{32} = 0.0625 \: moles \: of \: S}$


$\text{\underline{From\:the\:equation:}}$

$\boxed{Mg\:+\:S\:→\: MgS}$


It follows that one mole of Mg reacts with one mole of S.

We are given more mole of Mg than S, therefore S is the limiting reagent.


$\text{\underline{From\:equation,}}$

One mole of S gives one mole of MgS.

So, 0.0625 mole of S will react with 0.0625 mole of Mg to form 0.0625 mole of MgS.

Molar mass of MgS = $\sf\boxed{56\:g}$


$\text{\underline{Therefore,}}$


$\sf{\underline{Mass \: of \: MgS \: formed}}$

$\sf{ = 0.0625 \times 56.0 \: g}$

$\sf\boxed{ = 3.5 \: g \: of \: MgS}$


$\sf{\underline{Mass \: of \: Mg \: left \: unreacted}}$

$\sf{ = 0.0833 - 0.0625 \: mole \: of \: Mg}$

$\sf\boxed{ = 0.0208 \: mole \: of \: Mg}$


$\text{\underline{So,}}$

$\sf{Mass \: of \: Mg \: left \: unreacted}$

$\sf{ = Moles \: of \: Mg \times Molar \: mass \: of \: Mg}$

$\sf{ = 0.0208 \times 24 \: g \: of \: Mg}$

$\sf{ = 0.4992 \: g \: of \: Mg}$

$\sf\boxed{ = 0.5 \: g \: of \: Mg}$

Answer 2

Answer:

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