Đáp án:

 8. B

9. A

10. B

11. C

Giải thích các bước giải:

8. Oxit có công thức A2O3 => Oxit có hóa trị (III), loại D

$A_2O_3+6HCl\rightarrow 2ACl_3+3H_2O$

$n_{HCl}=0,96\left(mol\right)$

$TheoPT:n_{A_2O_3}=\frac{1}{6}n_{HCl}=0,16\left(mol\right)$

$\Rightarrow M_{A_2O_3}=2A+16.3=\frac{16,32}{0,16}=102$

$\Rightarrow A=27\left(Al\right)$

=> Chọn B

9. $n_{Fe_2O_3}=\frac{1,6}{160}=0,01\left(mol\right);n_{H_2SO_4}=\frac{100.10\%}{98}=0,102\left(mol\right)$

$Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O$

Lập tỉ lệ : $\frac{0,01}{1}<\frac{0,102}{3}\Rightarrow H_2SO_4dư$

Theo PT : $n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,01\left(mol\right)$

=> $m_{Fe_2\left(SO_4\right)_3}=0,01.400=4\left(g\right)$

=> Chọn A

10. Gọi x,y là số mol Mg, Fe

$Mg+2HCl\rightarrow MgCl_2+H_2$

$Fe+2HCl\rightarrow FeCl_2+H_2$

$n_{HCl}=3.0,2=0,6\left(mol\right)$

$\left \{ {{2x+2y=0,6} \atop {24x+56y=13,6}} \right.$

=> x=0,1 ; y=0,2

$m_{Fe}=0,2.56=11,2\left(g\right)$

=> Chọn B

11. $CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O$

$n_{CO_2}=\frac{3,36}{22,4}=0,15\left(mol\right)$

$TheoPT:n_{CaCO_3}=n_{CO_2}=0,15\left(mol\right)$

$\Rightarrow m_{CaCO_3}=0,15.100=15\left(g\right)$

=> Chọn C

Em tham khảo nha :

\(\begin{array}{l}
8)\\
B\\
{A_2}{O_3} + 6HCl \to 2AC{l_3} + 3{H_2}O\\
{n_{^{HCl}}} = 0,96 \times 1 = 0,96mol\\
{n_{{A_2}{O_3}}} = \dfrac{{{n_{HCl}}}}{6} = 0,16mol\\
{M_{{A_2}{O_3}}} = \dfrac{{16,32}}{{0,16}} = 102dvC\\
 \Rightarrow {M_A} = 27dvC\\
A:Al\\
CTHH:A{l_2}{O_3}\\
9)\\
A\\
{n_{F{e_2}{O_3}}} = \dfrac{{1,6}}{{160}} = 0,01mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{100 \times 10}}{{100}} = 10g\\
{n_{{H_2}S{O_4}}} = \dfrac{{10}}{{98}} = 0,102mol\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
\dfrac{{0,01}}{1} < \dfrac{{0,102}}{3} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{F{e_2}{{(S{O_4})}_3}}} = {n_{F{e_2}{O_3}}} = 0,01mol\\
{m_{F{e_2}{{(S{O_4})}_3}}} = 0,01 \times 400 = 4g\\
10)\\
B\\
{n_{HCl}} = 3 \times 0,2 = 0,6mol\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
hh:Fe(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
2a + 2b = 0,6\\
56a + 24b = 13,6
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,1\\
{m_{Fe}} = 0,2 \times 56 = 11,2g\\
11)\\
C\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,15mol\\
{m_{CaC{O_3}}} = 0,15 \times 100 = 15g
\end{array}\)