Đáp án D

\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

\(x\)                                                          \(x\)

\(MgCO_2+2HCl\rightarrow MgCl_2+H_2O+CO_2\uparrow\)

\(y\)                                                             \(y\)

Thu đc kết tủa \(BaSO_4\)

\(n_{BaSO_4}=\dfrac{1,7475}{233}=0,0075mol=n_{Ba\left(OH\right)_2dư}\)

\(\Rightarrow n_{Ba\left(OH\right)_2PƯ}=0,0225-0,0075=0,015mol=n_{BaCO_3}\)(BT Ba)

\(\Rightarrow n_{CO_2}=n_{BaCO_3}=0,015mol\left(BtC\right)\)

Ta có:

\(\left\{{}\begin{matrix}100x+84y=1,42\\x+y=n_{CO_2}=0,015\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,01\\y=0,005\end{matrix}\right.\)

\(m_{CaCO_3}=0,01\cdot100=1g\)

\(m_{MgCO_3}=0,005\cdot84=0,42g\)

a)

Gọi số mol Mg, Na₂CO₃ là a,b (mol)

=> 24a + 106.b = 13 (1)

\(n_{H_2}+n_{CO_2}=\dfrac{4,48}{33,4}=0,2\left(mol\right)\)

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

______a------>a------------>a------->a________(mol)

Na₂CO₃ + H₂SO₄ --> Na₂SO₄ + CO₂ + H₂O

__b---------->b----------->b------>b______________(mol)

=> a + b = 0,2 (2)

(1)(2) => \(\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%Mg=\dfrac{0,1.24}{13}100\%=18,46\%\\\%Na_2CO_3=100\%-18,46\%=81,54\%\end{matrix}\right.\)

b) 

PTHH: \(Ba\left(OH\right)_2+H_2SO_4->BaSO_4\downarrow+2H_2O\)

_________________k------------>k_______________(mol)

\(Ba\left(OH\right)_2+MgSO_4->BaSO_4\downarrow+Mg\left(OH\right)_2\downarrow\)

____________0,1---------->0,1---------->0,1_________(mol)

\(Ba\left(OH\right)_2+Na_2SO_4->BaSO_4\downarrow+2NaOH\)

____________0,1-------->0,1____________________(mol)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

_0,1---------->0,1______________________________(mol)

=> \(\left\{{}\begin{matrix}n_{BaSO_4}=k+0,2\\n_{MgO}=0,1\end{matrix}\right.\)

=> \(233.\left(k+0,2\right)+40.0,1=62,25\)

=> k = 0,05 (mol)

=> n_H2SO4 = 0,1 + 0,1 + 0,05 = 0,25 (mol)

=> \(V_{dd}=\dfrac{0,25}{1}=0,25\left(l\right)=250ml\)

Đáp án C

Chọn D

Dẫn hỗn hợp khí qua P b N O 3 2 thì H 2 S bị giữ lại; H 2 thoát ra khỏi dung dịch:

P b N O 3 2   +   H 2 S   →   P b S ( ↓   đ e n )   +   2 H N O 3

P b N O 3 2   +   H 2   →   k h ô n g   p h ả n   ứ n g

Dẫn khí thoát ra qua H 2 S O 4 đặc, thu được  H 2  khô (không lẫn hơi nước)