Đáp án:

$a/PTHH:CuO+H_2$$\longrightarrow$$Cu+H_2O$

$b/m_{Cu}=6,4g$

$c/V_{H_2}=2,24l$

Giải thích các bước giải:

 a/ $PTHH:CuO+H_2$$\longrightarrow$$Cu+H_2O$

b/

$n_{CuO}=\dfrac{m}{M}=$$\dfrac{8}{80}=0,1(mol)$

$PTHH:CuO+H_2$$\longrightarrow$$Cu+H_2O$

$\text{Tỉ lệ: 1 : 1 : 1 : 1 mol}$

$\Rightarrow$$n_{cu}=$$\dfrac{0,1.1}{1}=0,1(mol)$

$\Rightarrow$$m_{Cu}=n.M=0,1.64=6,4(g)$

$c/\Rightarrow$$n_{H_2}=$$\dfrac{0,1.1}{1}=0,1(mol)$

$\Rightarrow$$V_{H_2}=n.22,4=0,1.22,4=2,24(l)$

Đáp án + Giải thích các bước giải:

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

$b\big)$

Theo PT: $n_{CuO}=n_{Cu}=\frac{8}{80}=0,1(mol)$

$\to m-{Cu}=0,1.64=6,4(g)$

$c\big)$

Theo PT: $n_{H_2}=n_{Cu}=0,1(mol)$

$\to V_{H_2(đktc)}=0,1.22,4=2,24(l)$