1)

`n_(Cu)=(12,8)/64=0,2mol`

PTHH xảy ra:

`2Cu+O_2->2CuO`

Ta có: `n_(Cu)=n_(CuO)=0,2mol`

`m_(CuO)=0,2×(64+16)=16g`

b)`n_(O_2)=n_(Cu)/2=0,1mol`

`V_(O_2)=n×22,4=0,1×22,4=2,24l`

2)`n_(Zn)=16/65mol`

`Zn+2HCl->ZnCl_2+H_2`

`n_(H_2)=n_(Zn)=16/65mol`

`V_(H_2)=(16/65)×22,4=5,51l`

3)

a)`PTHH: Mg+2HCl->MgCl_2+H_2`

b)`n_(Mg)=(3,6)/24=0,15mol`

`n_(HCl)=(14,6)/(36,5)=0,4mol`

Lập tỉ lệ:

`n_(Mg)/1<n_(HCl)/2(0,15<0,2)`

Nên `Mg` phản ứng hết, `HCl` còn dư.

`n_(H_2)=n_(Mg)=0,15mol`

`V_(H_2)=0,15×22,4=3,36l`

Bài 1:

a) $n_{Cu}$= $\frac{12,8}{64}$=0,2 mol

2Cu + O2 --t0--> 2CuO

0,2       0,1                 0,2  

=> mCuO=0,2.80 =16 gam

b) $V_{O2}$=0,1.22,4=2,24 lít 

Bài 2: 

PTPƯ:   Zn+2HCl-->ZnCl2+H2

            $\frac{16}{65}$         $\frac{16}{65}$    

$n_{Zn}$= $\frac{16}{65}$ mol

b) $V_{H2}$=$\frac{16}{65}$.22,4=5,51 lít 

Bài 3:

a) Mg+2HCl -> MgCl2+H2 

b) nH2=nMg=3,6:24=0,15 mol

=> $V_{H2}$= 0,15.22,4=3,36 lít