Đáp án+Giải thích các bước giải:

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

$\text{A.}$

$\text{Ta có: $n_{Fe_{2}O_{3}}$=$\dfrac{68,9}{160}$= 0,430625 (mol)}$

$\text{Theo PT ta có: $n_{Fe}$=2$n_{Fe_{2}O_{3}}$=2 × 0,430625= 0,86125 (mol)}$

$\text{⇒ $m_{Fe}$=0,86125 × 56= 48,23 (g)}$

$\text{B.}$

$\text{Theo PT ta có: $n_{H_{2}}$=3$n_{Fe_{2}O_{3}}$=3 × 0,430625= 1,291875 (mol)}$

$\text{⇒ $V_{H_{2}}$=1,291875 × 22,4= 28,938 (lít)}$

$\text{C.}$

$\text{Theo PT ta có: $m_{H_{2}O}$=3$n_{Fe_{2}O_{3}}$=3 × 0,430625= 1,291875 (mol)}$

$\text{⇒ $m_{H_{2}O}$=1,291875 × 18= 23,25375 (g)}$

$#sad boy$

Đáp án:

a)

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

Ta có:

`n_{Fe_2O_3}=m/M=(68,9)/(160)= 0,430625 (mol)`

Theo PT: `n_{Fe}=2n_{Fe_2O_3}=2 xx 0,430625= 0,86125 (mol)`

`=> m_{Fe}=n xx M=0,86125 xx 56= 48,23 (g)`

b)

Theo PT: `n_{H_2}=3n_{Fe_2O_3}=3 xx 0,430625= 1,291875 (mol)`

`=> V_{H_2}=n xx 22,4=1,291875 xx 22,4= 28,938 (l)`

c)

Theo PT: `n_{H_2O}=3n_{Fe_2O_3}=3 xx 0,430625= 1,291875 (mol)`

`=> m_{H_2O}=n xx M=1,291875 xx 18= 23,25375 (g)`