$@Senseless$
$\rm PTHH:Al_2O_3+3H_2SO_4\xrightarrow{}Al_2(SO_4)_3+3H_2O$
$\rm n_{H_2SO_4}=\dfrac{49}{98}=0,5(mol)$
$\rm n_{Al_2O_3}=\dfrac{60}{102}=\dfrac{10}{17}(mol)$
Nhận xét : 
$\rm \dfrac{n_{H_2SO_4}}{3}<\dfrac{n_{Al_2O_3}}{1}$
$\rm \Rightarrow H_2SO_4$ hết và $\rm Al_2O_3$ dư

Theo phương trình : 
$\rm n_{Al_2O_3}=\dfrac{1}{3}n_{H_2SO_4}=\dfrac{0,5}{3}(mol)$
$\rm \Rightarrow m_{Al_2O_3(dư)}=(\dfrac{10}{17}-\dfrac{0,5}{3})\times102=43(g)$

Đáp án:

$\rm n_{H_2SO_4}=\dfrac{49}{98}=0,5\ (mol)\\n_{Al_2O_3}=\dfrac{60}{102}=\dfrac{10}{17}\ (mol)$

Tỉ lệ phản ứng: $\rm \dfrac{n_{Al_2O_3}}{1}>\dfrac{n_{H_2SO_4}}{3}\to Al_2O_3$ còn dư.

$\rm n_{Al_2O_3\ (du)}=\dfrac{10}{17}-\dfrac{1}{3}×0,5=\dfrac{43}{102}\ (mol)$

$\rm m_{Al_2O_3\ (du)}=\dfrac{43}{102}×102=43\ (g)$