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 Consider this reaction:       `Mg_2Si(s)+4H_2O(L) --> 2Mg(OH)_2(aq)+SiH_4(g)`...

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roshan-rox | Valedictorian

Posted August 10, 2013 at 5:48 PM via web

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 Consider this reaction:      

`Mg_2Si(s)+4H_2O(L) --> 2Mg(OH)_2(aq)+SiH_4(g)`

How many grams of excess reactant are left over if we start with 39.1 g of each reactant?

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted August 10, 2013 at 5:56 PM (Answer #1)

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Molar mass in `g/(mol)`

`Mg_2Si = 76.7`

`H_2O = 18`

Molar mass ratio

`Mg_2Si:H_2O = 76.7:18 = 4.26:1`

Mole ratio

`Mg_2Si:H_2O = 1:4`

Since the molar mass ratio is greater than molar ratio and we have constant mass from each substance the reactant with lower molar mass will remain as excess. Since `H_2O` has lower molar mass it will be the excess reactant.

`Mg_2Si` moles used for reaction `= 39.1/76.7 = 0.509`

`H_2O ` required for the reaction `= 0.509x4 = 2.039`

`H_2O` available for the reaction `= 39.1/18 = 2.172`

Remaining `H_2O = 2.172-2.039 = 0.133 mols`

Remaining `H_2O = 0.133xx18 = 2.398g`

So 2.398g of `H_2O` will be remain as excess.

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