Đáp án:

 b) 3,7185l

c) 5,6g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{{9,75}}{{65}} = 0,15\,mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,15\,mol\\
{V_{{H_2}}} = 0,15 \times 24,79 = 3,7185l\\
c)\\
F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
{n_{Fe}} = 0,15 \times \dfrac{2}{3} = 0,1\,mol\\
{m_{Fe}} = 0,1 \times 56 = 5,6g
\end{array}\)

$\text{Đáp án+Giải thích các bước giải:}$

$\text{A.}$

$\text{Zn + 2HCl→$ZnCl_{2}$ + $H_{2}$ (1)}$

$\text{B.}$

$\text{Ta có: $n_{Zn}$=$\dfrac{9,75}{65}$= 0,15 (mol)}$

$\text{Theo PT (1) ta có: $n_{H_{2}}$=$n_{Zn}$= 0,15 (mol)}$

$\text{⇒ $V_{H_{2}}$=0,15 × 24,79= 3,7185 (lít)}$

$\text{C.}$

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

$\text{Theo PT (2) ta có: $n_{Fe}$=$\dfrac{2}{3}$$n_{H_{2}}$=$\dfrac{2}{3}$ × 0,15= 0,1 (mol)}$

$\text{⇒ $m_{Fe}$=0,1 × 56= 5,6 (g)}$