Đáp án + Giải thích các bước giải:

Câu 13:

$a)$

$n_{Fe_2O_3}=\frac{32}{160}=0,2(mol)$

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

Theo PT: $n_{Fe}=2n_{Fe_2O_3}=0,4(mol)$

$\to m_{Fe}=0,4.56=22,4(g)$

$b)$

Theo PT: $n_{H_2}=3n_{Fe_2O_3}=0,6(mol)$

$\to V_{H_2(đktc)}=0,6.22,4=13,44(l)$

Câu 15:

$a)Zn+2HCl\to ZnCl_2+H_2$

$b)$

$n_{Zn}=\frac{13}{65}=0,2(mol)$

Theo PT: $n_{H_2}=n_{Zn}=0,2(mol)$

$\to V_{H_2(đktc)}=0,2.22,4=4,48(l)$

$c)$

Đổi $1680ml=1,68(l)$

$n_{O_2}=\frac{1,68}{22,4}=0,075(mol)$

$2Zn+O_2\xrightarrow{t^o}2ZnO$

Vì $\frac{n_{Zn}}{2}>n_{O_2}\to Zn$ dư

Theo PT: $n_{ZnO}=0,15(mol);n_{Zn(dư)}=0,2-0,15=0,05(mol)$

$\to m_{CR}=0,15.81+0,05.65=15,4(g)$

Em tham khảo nha :

\(\begin{array}{l}
13)\\
a)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = \dfrac{{32}}{{160}} = 0,2mol\\
{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,4mol\\
{m_{Fe}} = 0,4 \times 56 = 22,4g\\
b)\\
{n_{{H_2}}} = 3{n_{F{e_2}{O_3}}} = 0,6mol\\
{V_{{H_2}}} = 0,6 \times 22,4 = 13,44l\\
15)\\
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{{13}}{{65}} = 0,2mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
c)\\
2Zn + {O_2} \to 2ZnO\\
{n_{{O_2}}} = \dfrac{{1,68}}{{22,4}} = 0,075mol\\
\dfrac{{0,2}}{2} > \dfrac{{0,075}}{1} \Rightarrow Zn\text{ dư}\\
{n_{Z{n_d}}} = 0,2 - 0,075 \times 2 = 0,05mol\\
{n_{ZnO}} = 2{n_{{O_2}}} = 0,15mol\\
m = {m_{Z{n_d}}} + {m_{ZnO}} = 0,05 \times 65 + 0,15 \times 81 = 15,4g
\end{array}\)