Em tham kharo nha :

\(\begin{array}{l}
3)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{{4,8}}{{24}} = 0,2mol\\
{n_{{H_2}}} = {n_{Mg}} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
FeO + {H_2} \to Fe + {H_2}O\\
{n_{FeO}} = {n_{{H_2}}} = 0,2mol\\
{m_{FeO}} = 0,2 \times 72 = 14,4g\\
{n_{Fe}} = {n_{{H_2}}} = 0,2mol\\
{m_{Fe}} = 0,2 \times 56 = 11,2g\\
4)\\
a)\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
3Fe + 2{O_2} \to F{e_3}{O_4}\\
{n_{Al}} = \dfrac{{27,8 \times 19,2\% }}{{27}} = 0,2mol\\
{n_{Fe}} = \dfrac{{27,8 - 0,2 \times 27}}{{56}} = 0,4mol\\
{n_{{O_2}}} = \dfrac{3}{4}{n_{Al}} + \dfrac{2}{3}{n_{Fe}} = \dfrac{5}{{12}}mol\\
{V_{{O_2}}} = \dfrac{5}{{12}} \times 22,4 = 9,333l\\
{V_{kk}} = 5{V_{{O_2}}} = 46,67l\\
b)\\
{m_{{O_2}}} = \frac{5}{{12}} \times 32 = 13,33g\\
BTKL:\\
{m_{kl}} + {m_{{O_2}}} = {m_{hh}}\\
 \Rightarrow {m_{hh}} = 27,8 + 13,33 = 41,13g
\end{array}\)

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

$\text{Bài 3:}$

$\text{Mg + 2HCl→$MgCl_{2}$ + $H_{2}$ (1)}$

$\text{A.}$

$\text{Ta có: $n_{Mg}$=$\dfrac{4,8}{24}$= 0,2 (mol)}$

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

$\text{B.}$

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

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

$\text{⇒ $m_{FeO}$=0,2.72= 14,4 (g)}$

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

$\text{⇒ $m_{Fe}$=0,2.56= 11,2 (g)}$

$\text{Câu 4:}$

$\text{4Al + $3O_{2}$$\xrightarrow{t^o}$$2Al_{2}O_{3}$ (1)}$

$\text{3Fe + $2O_{2}$$\xrightarrow{t^o}$$Fe_{3}O_{4}$ (2)}$

$\text{A.}$

$\text{Ta có: $m_{Al}$=$\dfrac{19,2.27,8}{100}$= 5,4 (g)}$

$\text{⇒ $n_{Al}$=$\dfrac{5,4}{27}$= 0,2 (mol)}$

$\text{$m_{Fe}$=27,8 - 5,4= 22,4 (g)}$

$\text{⇒ $n_{Fe}$=$\dfrac{22,4}{56}$= 0,4 (mol)}$

$\text{Theo PT (1): $n_{O_{2}}$=$\dfrac{3}{4}$$n_{Al}$=$\dfrac{3}{4}$.0,2= 0,15 (mol)}$

$\text{Theo PT (2): $n_{O_{2}}$=$\dfrac{2}{3}$$n_{Fe}$=$\dfrac{2}{3}$.0,4= $\dfrac{4}{15}$ (mol)}$

$\text{⇒ ∑$n_{O_{2}}$=0,15 + $\dfrac{4}{15}$= $\dfrac{5}{12}$ (mol)}$

$\text{⇒ $V_{KK}$=5.($\dfrac{5}{12}$.22,4)≈ 46,67}$ `(l)`

$\text{B.}$

$\text{$m_{O_{2}}$=$\dfrac{5}{12}$.32≈ 13,33 (g)}$

$\text{Theo ĐLBTKL, ta có:}$

$\text{$m_{HH}$ + $m_{O_{2}}$ = $m_{HH chất rắn}$}$

$\text{⇔ 27,8 + 13,33 = $m_{HH chất rắn}$}$

$\text{⇒ $m_{HH chất rắn}$=27,8 + 13,33= 41,13 (g)}$