Answer
a) $Cr_2O_3$ will be left over.
b) $m(Cr_2O_3) = 15.124g$
c) $m(Cr) = 18.876g$
Work Step by Step
Let's consider the reaction that takes place:
$Cr_2O_3 + 2Al\rightarrow 2Cr + Al_2O_3$
Since we are given the mass of both reagents, first we need to determine which one is the limiting reagent. We can calculate the amounts of both of reagents:
$n(Cr_2O_3) = \frac{m(Cr_2O_3)}{M(Cr_2O_3)} = \frac{42.7g}{152\frac{g}{mol}} = 0.281mol$
$n(Al) = \frac{m(Al)}{A(Al)} = \frac{9.8g}{27\frac{g}{mol}} = 0.363mol$
a) Observing stoichiometric coefficients, we can conclude that the amount of chromium(III) oxide required for the reaction is half the amount of aluminum, which means that for the reaction of $0.363mol$ of $Al$, $0.1815mol$ of $Cr_2O_3$ is required. Therefore, chromium(III) oxide will be left over.
b) Since $0.1815mol$ of $Cr_2O_3$ is required for the reaction, the excess amount of $Cr_2O_3$ is equal to $0.281mol - 0.1815mol = 0.0995mol$, which corresponds to the following mass:
$m(Cr_2O_3) = n(Cr_2O_3)\times M(Cr_2O_3) = 0.0995mol \times 152\frac{g}{mol} = 15.124g$
c) The amount of chromium produced in this reaction equals to the amount of aluminum that has reacted, hence the amount of the obtained chromium is equal to $0.363mol$. That corresponds to the following mass:
$m(Cr) = n(Cr)\times A(Cr) = 0.363mol \times 52\frac{g}{mol} = 18.876g$
