Part $2$: Limiting reagents and percent Yield

Use the balanced equation below to answer the following equations.

$2Al+3$ZnO$A{l}_2{O}_3+3Zn$

If $55\mathrm{.}0$ grams of aluminum reacts with $90\mathrm{.}0$ grams of zinc oxide, calloulate

the following

a$)$ What is the limiting reactant? What is the excess reactant? $(10$ pts$)$

Please show calculations to justify your answer

$\frac{{55\mathrm{.}0}_{g}Al}{{54\mathrm{.}0}_{\frac{g}{1m(m)}}^{1}Al}=1\mathrm{.}0m,\frac{90(g)2nO}{3(9\mathrm{.}4)1\frac{g}{mm}2(2n0)}$

$ER$

$65\mathrm{.}4+16=81\mathrm{.}4$

LR

b$)$ How much of the excess reactant is left over? $(6$ pt$)$

gof Al lept

$\{\begin{array}{c}\text{:}=19\mathrm{.}9\text{g}\text{A}\text{l}\end{array}$

Did I do parts a & b correctly? If so$,$ how do I calculate the $\%$ yield??