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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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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...

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