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How much magnesium sulphide can be obtai...

How much magnesium sulphide can be obtained from `2.00 g` of `Mg` and `2.00 g` of `S` by the reaction.
`Mg + S rarr MgS`. Which is the limiting reagent? Calculate the amount of one of the reactants which remains unreacted?

Text Solution

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`Mg + S rarr MgS`
`24 g 32 g 32 + 24 = 56 g`
`32 g` of `S` in obtianed from `24 g` of `Mg`
`2 g` of `S` in obtained from `= (24)/(32) xx 2 = 1.5 g of Mg`
So, `S` is completely consumed and `Mg` is left.
Hence, `S` is the limiting reagent.
Therefore, the amount of product, i.e., `MgS`, will be determined from `S` and not from `Mg`.
`32 g` of `S = 56 g of MgS`
`2 g of S = (56)/(32) xx 2 = 3.5 g of MgS`
Amount of `Mg` unreacted `= 2 - 1.5 = 0.5 g`
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