$+)$

$2K+2H_2O \to 2KOH+H_2$

$BaO+H_2O \to Ba(OH)_2$

$SO_2+H_2O \to H_2SO_3$

$+)$

`n_(Na)=(17,25)/23=0,75(mol)`

$a)PTHH :$

$2Na+2H_2O \to 2NaOH+H_2$

$b)$

Theo `PT,n_(NaOH) =n_(Na)=0,75(mol)`

`=>m=m_(NaOH)=0,75.40=30(g)`

Theo `PT,n_(H_2)=1/2n_(Na)=0,375(mol)`

 `=>V=V_(H_2)=0,375.22,4=8,4(l)`

$c)$

$PT:H_2+CuO\xrightarrow{t^0}Cu+H_2O$

Theo `PT,n_(Cu)=n_(H_2)=0,375(mol)`

`=>m_(Cu)=0,375.64=24(g)`

Đáp án + Giải thích các bước giải:

$2K+2H_2O\to 2KOH+H_2$

$BaO+H_2O\to Ba(OH)_2$

$SO_2+H_2O\leftrightharpoons H_2SO_3$

$$---------------$$

$a\big)2Na+2H_2O\to 2NaOH+H_2$

$b\big)$

$n_{Na}=\frac{17,25}{23}=0,75(mol)$

Theo PT: $n_{NaOH}=n_{Na}=0,75(mol);n_{H_2}=\frac{1}{2}n_{Na}=0,375(mol)$

$\to m=m_{NaOH}=0,75.40=30(g);V=V_{H_2}=0,375.22,4=8,4(l)$

$c\big)$

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

Theo PT: $n_{Cu}=n_{H_2}=0,375(mol)$

$\to m_{Cu}=0,375.64=24(g)$