Đáp án + Giải thích các bước giải:

$c\big)$

$n_{Fe_2O_3}=\frac{150.80\%}{160}=0,75(mol)$

$n_{CuO}=\frac{150-160.0,75}{80}=0,375(mol)$

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

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

Theo PT: $\sum n_{H_2}=3n_{Fe_2O_3}+n_{CuO}=2,625(mol)$

$Zn+2HCl\to ZnCl_2+H_2$

Theo PT: $n_{Zn}=n_{H_2}=2,625(mol)$

$\to m_{Zn}=2,625.65=170,625(g)$

Đáp án+Giải thích các bước giải:}$

$\text{CuO + $H_{2}$$\xrightarrow{t^o}$Cu + $H_{2}O$ (1)}$

$\text{$Fe_{2}O_{3}$ + $3H_{2}$$\xrightarrow{t^o}$2Fe + $3H_{2}O$ (2)}$

$\text{Zn + 2HCl→$ZnCl_{2}$ + $H_{2}$ (3)}$

$\text{A.}$

$\text{$m_{Fe_{2}O_{3}}$=$\dfrac{80 × 150}{100}$= 120 (g)}$

$\text{$m_{CuO}$=150 - 120= 30 (g)}$

$\text{B.}$ 

$\text{Ta có: $n_{Fe_{2}O_{3}}$=$\dfrac{120}{160}$= 0,75 (mol)}$

$\text{$n_{CuO}$=$\dfrac{30}{80}$= 0,375 (mol)}$

$\text{Theo PT (1): $n_{H_{2}}$=$n_{CuO}$= 0,375 (mol)}$

$\text{Theo PT (2): $n_{H_{2}}$=3$n_{Fe_{2}O_{3}}$=3 × 0,75= 2,25 (mol)}$

$\text{⇒ ∑$n_{H_{2}}$=0,375 + 2,25= 2,625 (mol)}$

$\text{⇒ $V_{H_{2}}$=2,625 × 22,4= 58,8 (lít)}$

$\text{C.}$

$\text{Theo PT (3): $n_{Zn}$=$n_{H_{2}}$= 2,625 (mol)}$

$\text{⇒ $m_{Zn}$=2,625 × 65= 170,625 (g)}$