Đáp án:

 

Giải thích các bước giải:

 Bài 1

a) $3H2+Fe2O3--to--->2Fe+3H2O$

b)$2Al+6HCl--->2AlCl3+3H2$

c)$H2+PbO--to-->Pb+H2O$

d)$2H2+O2--to>2H2O$

E)$Fe+H2SO4-----to--FeSO4+H2$

Bài 2

$a)Zn+2HSO4--->ZnSO4+H2$

b)

$nH2SO4=49/98=0,5(mol)$

$nZn=nH2SO4=0,5(mol)$

=>$mZn=0,5.65=32,5(g)$

c)

$nH2=$nH2SO4=0,5(mol)$

=>$VH2=0,5.22,4=11,2(l)$

Câu 3

a)

$2Al+6HCl-->2AlCl3+3H2$

x------------------------1,5x(mol)$

$Mg+2HCl-->MgCl2+H2$

y-----------------------y(mol)$

b)

$nH2=8,96/22,4=0,4(mol)$

Theo bài ta có hpt

$27x+24y=7,8$

$1,5x+y=0,4$

=>$x=0,2,y=0,1$

=>%$mAl=0,2.27/7,8.100$%$=69,23$%

=>%$mMg=100-69,23=30,77$%

1/

$3H_2+Fe_2O_3\overset{t^o}\to 2Fe+3H_2O$

$2Al+6HCl\to 2AlCl_3+3H_2$

$H_2+PbO \to Pb+H_2O$

$2H_2+O_2\overset{t^o}\to 2H_2O$

$Fe+H_2SO_4\to FeSO_4+H_2$

2/

$n_{H_2SO_4}=49/98=0,5mol$

$a/Zn+H_2SO_4\to ZnSO_4+H_2$

b/Theo pt :

$n_{Zn}=n_{H_2SO_4}=0,5mol$

$⇒m_{Zn}=0,5.65=32,5g$

c/Theo pt :

$n_{H_2}=n_{H_2SO_4}=0,5mol$

$⇒V_{H_2}=0,5.22,4=11,2l$

3/

$n_{H_2}=8,96/22.4=0,4mol$

$a/2Al+6HCl\to 2AlCl_3+3H_2$

$Mg+2HCl\to MgCl_2+H_2$

Gọi $n_{Al}=a ; n_{Mg}=b$

b/Ta có :

$m_{hh}=27a+24b=7,8$

$n_{H_2}=1,5a+b=0,4$

$⇒\left\{\begin{matrix}
27a+24b=7,8 & \\ 
1,5a+b=0,4 & 
\end{matrix}\right.$

$\left\{\begin{matrix}
a=0,2 & \\ 
b=0,1 & 
\end{matrix}\right.$

$⇒m_{Al}=0,2.27=5,4g$

$⇒\%m_{Al}=\dfrac{5,4.100\%}{7,8}=69,23\%$

$⇒\%m_{Mg}=100\%-69,23\%=30,77\%$