a)

$R + 2HCl \to RCl_2 + H_2$
$2B + 6HCl \to 2BCl_3 + 3H_2$

$n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$
$n_{HCl} = 2n_{H_2} = 0,5(mol)$

Bảo toàn khối lượng : $m_{muối} = m_{kl} + m_{HCl} - m_{H_2}$

$= 9,2 + 0,5.36,5 - 0,25.2 = 26,95(gam)$

b) $V_{dd\ HCl} = \dfrac{0,5}{2} = 0,25(lít)$

1. C

2. A

3. D

Chất rắn không tan là Cu.

⇒ m_Cu = 9 (g)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\) ⇒ 27x + 56y = 20 - 9 (1)

Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{3}{2}x+y=0,4\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{9}{20}.100\%=45\%\\\%m_{Al}=\dfrac{0,2.27}{20}.100\%=27\%\\\%m_{Fe}=28\%\end{matrix}\right.\)

Ta có: \(n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)

PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

Theo PT: \(n_{Mg}=n_{H_2}=0,075\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,075.24}{8}.100\%=22,5\%\\\%m_{Ag}=77,5\%\end{matrix}\right.\)

Mg+H2SO4->MgSO4+H2

Ag ko tác dụng với H2SO4

nH2=1,68/22,4=0,07(mol)

=>nMg=0,07(mol)

=>mMg=0,07*24=1,68(g)

=>nAg=8-1,68=6,32(g)

%Mg=1,68/8=21%

%Ag=100%-21%=79%

\(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\)

PTHH: 2Na + 2H₂O ---> 2NaOH + H₂ 

             2<-------------------2

=> m_Na = 2.23 = 46 (g); m_Cu = 100 - 46 = 54 (g)

\(Cu+2H_2O\rightarrow Cu\left(OH\right)_2+H_2\)

\(1:2:1:1\left(mol\right)\)

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{22,4}{22,4}=1\left(mol\right)\)

\(m_{Cu}=n.M=1.64=64\left(g\right)\)

\(m_{Na}=100-64=36\left(g\right)\)

\(S+O_2\underrightarrow{t^o}SO_2\)

\(1:1:1\left(mol\right)\)

\(0,08:0,08:0,08\left(mol\right)\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{22,4}{22,4}=1\left(mol\right)\)

\(n_{SO_2}=\dfrac{m}{M}=\dfrac{5,12}{64}=0,08\left(mol\right)\)

\(m_S=n.M=0,08.32=2,56\left(g\right)\)

\(n_{O_2}=n_{SO_2}=0,08\left(mol\right)\)

\(n_{O_2\left(dư\right)}=1-0,08=0,92\left(mol\right)\)

\(m_{O_2\left(dư\right)}=n.M=0,92.32=29,44\left(g\right)\)

cho hỏi là 0,08 mol là lấy đâu ra z

Đặt tổng số hạt p, n, e của A và B lần lượt là p, n, e

`=>` \(\left\{{}\begin{matrix}p+e+n=78\\p+e-n=26\\p=e\end{matrix}\right.\Leftrightarrow p=e=n=26\)

`=>` \(\left\{{}\begin{matrix}p_A+p_B=26\\2p_A-2p_B=28\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p_A=20\\p_B=6\end{matrix}\right.\)

Vậy A là Canxi (Ca); B là Cacbon (C)

a) Sửa đề: K₂PO₄ -> K₃PO₄ 

`K_3PO_4 + 3HCl -> 3KCl + H_3PO_4`

`HCl + NaHCO_3 -> NaCl + CO_2 + H_2O`

`2HCl + K_2CO_3 -> 2KCl + CO_2 + H_2O`

b)

`2KHCO_3 + H_2SO_4 -> K_2SO_4 + 2CO_2 + 2H_2O`

`H_2SO_4 + Na_2CO_3 -> Na_2SO_4 + CO_2 + H_2O`

`3H_2SO_4 + Mg_3(PO_4)_2 -> 3MgSO_4 + 2H_3PO_4`

`H_2SO_4 + CaCO_3 -> CaSO_4 + CO_2 + H_2O`

`2NaOH + CO_2 -> Na_2CO_3 + H_2O`

`BaCl_2 + Na_2SO_4 -> 2NaCl + BaSO_4`

`MgCl_2 + K_2CO_3 -> MgCO_3 + 2KCl`

\(2KI+O_3+H_2O\rightarrow2KOH+O_2+I_2\\ 10FeSO_4+2KMnO_4+8H_2SO_4\rightarrow5Fe_2\left(SO_4\right)_3+K_2SO_4+2MnSO_4+8H_2O\)