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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? |
Answer» (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)` |
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