Bài 1:

$a,PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ b,n_{H_2}=\dfrac{37,185}{24,79}=1,5(mol)\\ \Rightarrow m_{H_2}=1,5.2=3(g)\\ \text{Bảo toàn KL: }m_{Al}+m_{H_2SO_4}=m_{Al_2(SO_4)_3}+m_{H_2}\\ \Rightarrow m_{H_2SO_4}=3+171-27=147(g)$

Bài 2:

$n_{O_2}=\dfrac{2,479}{24,79}=0,1(mol)\\ \Rightarrow m_{O_2}=0,1.32=3,2(g)\\ a,PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ b,\text{Bảo toàn KL: }m_{Fe}+m_{O_2}=m_{Fe_3O_4}\\ \Rightarrow m_{Fe_3O_4}=3,2+8,4=11,6(g)$

Bài 3:

$\text{Bảo toàn KL: }m_{Mg}+m_{O_2}=m_{MgO}\\ \Rightarrow m_{O_2}=m_{MgO}-m_{Mg}=15-9=6(g)$

Bài 4:

$\text{Bảo toàn KL: }m_{KClO_3}=m_{KCl}+m_{O_2}\\ \Rightarrow m_{KCl}=m_{KClO_3}-m_{O_2}=24,5-9,6=14,9(g)$