$\text {PTHH:}$$\text {$ 4Al+3O2 \xrightarrow{{ t^o}}$2Al2O3}$
$\text {·$n_{Al}$= $\frac{54}{27}$ =2 (mol) }$
$\text {·Theo pt: $n_{Al2O3}$= $\frac{1}{2}$$n_{Al}$ }$
$\text {=$\frac{1}{2}$. 2 =1 (mol) }$
$\text {⇒$m_{Al2O3}$= 1. 102 =102 (g) }$
$\text {Vậy $m_{Al2O3}$ =102 g }$
$\text {Chúc bạn học tốt~}$
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