In the reaction: `2Al + Cr_(2) O_(3) rarr Al_(2) O_(3) + 2 Cr`,
`4.98 g` of `Al` reacted with `20.0 g Cr_(2)O_(3)`. How much grams of reactant ramains at the completion of the reaction?

Explanation

(Atomic weight of `Al` and `Cr = 27` and 52, `Mw` of `Cr_(2)O_(3) = 152)`
Moles of `Al = (4.98)/(27 Al) = 0.184 "mol"`
`= (0.184)/(2) = 0.92 "mol of" Cr_(2) O_(3)`
Since 2 mol `Al` is required for 1 mol of `Cr_(2)O_(3)`.
So, `Al` is the limiting reagent and `Cr_(2)O_(3)` is excess.
Moles of `Cr_(2)O_(3)` in excess `= (0.131 - 0.092) = 0.039`
`~~ 0.04 "moo"`
Weight of `Cr_(2) O_(3)` is excess `= 0.04 xx 150 ~~ 6 g Cr_(2) O_(3)`