Em tham khảo nha :

\(\begin{array}{l}
1)\\
C\\
2)\\
D\\
\% Ca = \dfrac{{40}}{{100}} \times 100\%  = 40\% \\
3)\\
A\\
F{e_x}{O_y} + y{H_2} \to xFe + y{H_2}O\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_A} = \dfrac{{{n_{{H_2}}}}}{y} = \dfrac{{0,2}}{y}mol\\
{M_A} = 11,6:\dfrac{{0,2}}{y} = 58y\\
 \Rightarrow 56x + 16y = 58y\\
 \Rightarrow 56x = 42y\\
 \Rightarrow \dfrac{x}{y} = \dfrac{3}{4}\\
 \Rightarrow CTHH:F{e_3}{O_4}\\
4)\\
D\\
Ca{(OH)_2}\\
5)\\
B\\
6)\\
A\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
B1:\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4}\\
2{C_4}{H_{10}} + 13{O_2} \to 8C{O_2} + 10{H_2}O\\
B2:\\
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
b)\\
{n_{Al}} = \dfrac{{1,62}}{{27}} = 0,06mol\\
{n_{HCl}} = 3{n_{Al}} = 0,18mol\\
{m_{HCl}} = 0,18 \times 36,5 = 6,57g\\
{m_{ddHCl}} = \dfrac{{6,57 \times 100}}{{10}} = 65,7g\\
c)\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,06mol\\
{m_{AlC{l_3}}} = 0,06 \times 133,5 = 8,01g\\
{m_{ddspu}} = 1,62 + 65,7 - \dfrac{3}{2} \times 0,06 \times 2 = 67,14g\\
C{\% _{AlC{l_3}}} = \dfrac{{8,01}}{{67,14}} \times 100\%  = 11,9\% 
\end{array}\)

1 C

2D
3A

4D

5B

6A

Tự luận

 BÀI 1

1,Fe₂O₃ + 3H₂  =>  2Fe  + 3H₂

2,Mg + 2HCl  => MgCl₂  + H₂

3,P₂ O₅  +2H₂O  => 2H₃PO₄

4,C₄H₁₀ +6O₂ => 4CO₂   + 5H₂O

BÀI 2 Theo đầu bài ta có n_{Al = 0,06 (mol)}

a, Al + 2HCl => AlCl_{3  +}H₂

    1            2                1          1      (mol)

   0,06        0,12          0,06      0,06     (mol)

_{mHCl= 0,12 .36,5 = 4,392 (g)}

_{mà ta có 10% HCl  = 4,38 . 10 = 43,92 (g)}

_{=>m ct = 1,62 (g)}

_{=> m dd = 6,012(g)}

_{C% = 1,62/6,012 x 100%= 26,946}