Đáp án:

$a,$ $V_{H_2}$  = $1,12(l)$

$b,$ $mHCl$  = $3,65(g)$

$c,$ $mCu$  = $3,2(g)$

Giải thích các bước giải:

$nFe$ = $\frac{m}{M}$ = $\frac{2,8}{56}$ = $0,05(mol)$

PTHH: $\rm Fe + 2HCl → FeCl_2 + H_2$

$\text{0,05(mol)→0,1(mol)        →0,05(mol)}$

$a,$

$V_{H_2}$ = $n * 22,4 = 0,05 * 22,4 = 1,12(l)$

$b,$

$mHCl$ = $n * M = 0,1 * 36,5 = 3,65(g)$

$c,$

PTHH:

$H_2$ + $CuO$ $\xrightarrow[]{t^o}$ $H_2O$ + $Cu$

-Theo PT: $nCu$ = $nH_2$ = $0,05(mol)$

$→$ $mCu$ = $0,05 * 64 = 3,2(g)$

$a)$

`n_(Fe)=(2,8)/56=0,05(mol)`

$PT:Fe+2HCl \to FeCl_2+H_2$

Theo `PT,n_(H_2)=n_(Fe)=0,05(mol)`

`=>V_(H_2)(đktc)=0,05.22,4=1,12(l)`

$b)$

Theo `PT,n_(HCl)=2n_(Fe)=0,1(mol)`

`=>m_(HCl)=0,1.36,5=3,65(g)`

$c)$

$PT:CuO+H_2\xrightarrow{t^o}Cu+H_2O$

Theo `PT,n_(Cu)=n_(H_2)=0,05(mol)`

`=>m_(Cu)=0,05.64=3,2(g)`