Đáp án:

 3. b. \(\%m_{Zn}=80\%; \%m_{Al}=20\%\)

c. V = 10,64 lít

4. m = 8,98 gam

\(\text C_{M\ Ca(OH)_2}=0,12\ \text M\)

Giải thích các bước giải:

 Câu 3.

a. Gọi số mol Zn, Al trong 20,3 gam X lần lượt là a,b

\(⇒65a+27b=20,3\ \text{gam}\)

\(Zn+H_2SO_4\to ZnSO_4+H_2\\ 2Al+3H_2SO_4\to Al_2{(SO_4)}_3+3H_2\)

b. \(m_{\text{muối khan}}=161a+171b=65,9\ \text{gam}\)

\(⇒\begin{cases}a=0,25\ \text{mol}\\\\ b=0,15\ \text{mol}\end{cases}\)

\(⇒m_{Zn}=65\cdot 0,25=16,25\ \text{gam}\)

\(⇒\%m_{Zn}=\dfrac{16,25}{20,3}\cdot 100\%=80\%⇒\%m_{Al}=20\%\)

c. \(m_{SO_4^{2-}}=65,9-20,3=45,6\ \text{gam}⇒ n_{SO_4^{2-}}=\dfrac{45,6}{96}=0,475\ \text{mol}⇒ n_{H_2}=0,475\ \text{mol}⇒ V_{H_2}=0,475\cdot 22,4=10,64\ \text{lít}\)

Câu 4. 

\(MgCO_3\xrightarrow{t^\circ} MgO+ CO_2\ (1)\\ BaCO_3\xrightarrow{t^\circ} BaO+CO_2\ (2)\\ CO_2+Ca(OH)_2\to CaCO_3+H_2O\ (3)\\ 2CO_2+Ca(OH)_2\to Ca{(HCO_3)}_2\ (4)\\ Ca{(HCO_3)}_2\xrightarrow{t^\circ} CaCO_3+H_2O+ CO_2\ (5)\\ n_{CaCO_3 \ (5)}=\dfrac{2}{100}=0,02\ \text{mol}\)

\(⇒ n_{Ca{(HCO_3)}_2}=n_{CaCO_3\ (5)}=0,02\ \text{mol}\)

\(⇒ n_{CO_2\ (4)}=2\cdot n_{Ca{(HCO_3)}_2}=2\cdot 0,02=0,04\ \text{mol}\)

\(n_{CaCO_3\ (3)}=\dfrac{4}{100}=0,04\ \text{mol} ⇒ n_{CO_2\ (3)}=n_{CaCO_3\ (3)}=0,04\ \text{mol}\)

\(⇒\sum n_{CO_2}=0,04+0,04=0,08\ \text{mol}\)

\(⇒\sum m_{CO_2}=0,08\cdot 44=3,52\ \text{gam}\)

BTKL: \(⇒ m=3,52+5,46=8,98\ \text{gam}\)

Theo PTHH \((3) ⇒ n_{Ca(OH)_2\ (3)}=n_{CO_2\ (3)}=0,04\ \text{mol}\)

Theo PTHH \((4) ⇒ n_{Ca(OH)_2\ (4)}=\dfrac 12\cdot n_{CO_2\ (4)}=\dfrac 12\cdot 0,04=0,02\ \text{mol}\)

\(⇒\sum n_{Ca(OH)_2}=0,04+0,02=0,06\ \text{mol}\)

\(⇒ \text C_{M\ Ca(OH)_2}=\dfrac{0,06}{0,5}=0,12\ \text M\)