1/

$PTPƯ:$

$1/2SO_2+O_2\xrightarrow{t^o,V_2O_5} 2SO_3$

$2/SO_3+H_2O\xrightarrow{} H_2SO_4$

$3/H_2SO_4+2K\xrightarrow{} K_2SO_4+H_2↑$

$4/K_2SO_4+BaCl_2\xrightarrow{} BaSO_4↓+2KCl$

2/

$a,PTPƯ:2Al+3H_2SO_4\xrightarrow{} Al_2(SO_4)_3+3H_2↑$

$b,n_{H_2}=\dfrac{13,44}{22,4}=0,6mol.$

Chất rắn không tan là $Cu$

$Theo$ $pt:$ $n_{Al}=\dfrac{2}{3}n_{H_2}=0,4mol.$

$⇒m_{Al}=0,4.27=10,8g.$

$⇒m=m_{Al}+m_{Cu}=10,8+4,6=15,4g.$

$c,\%m_{Al}=\dfrac{10,8}{15,4}.100\%=70,13\%$

$\%m_{Cu}=\dfrac{4,6}{15,4}.100\%=29,87\%$

chúc bạn học tốt!

2SO2+O2=>2SO3

SO3+H2O=>H2SO4

H2SO4+2K=>K2SO4+H2

K2SO4+Ba=>BaSO4+2K

câu 2:

a,2Al+3H2SO4=>Al2(SO4)3+3H2

Cu không tác dụng

b,ta có vH2=13.44(lít)=>nH2=$\frac{13.44}{22.4}$=0.6(mol) 

=>nAl=$\frac{0.6*2}{3}$=0.4(mol) 

=>mAl=0.4*24=10.8(g)

=>mX=10.8+4.6=15.4(g)

c,%Al=$\frac{10.8}{15.4}$*100=70.129% 

=>%Cu=100-70.129=29.871%