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?
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`
`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.