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PTHH : \(CaS+2HBr-->CaBr_2+H_2S\uparrow\)

\(n_{H_2S}=\frac{0,672}{22,4}=0,03\left(mol\right)\)

Theo pthh : \(n_{CaS}=n_{H_2S}=0,03\left(mol\right)\)

                   \(n_{HBr}=2n_{H_2S}=0,06\left(mol\right)\)

                   \(n_{CaBr_2}=n_{H_2S}=0,03\left(mol\right)\)

\(\Rightarrow\hept{\begin{cases}m_{CaS}=0,03\cdot72=2,16\left(g\right)=m\\m_{ddHBr}=\frac{0,06\cdot81}{9,72}\cdot100=50\left(g\right)=m_1\end{cases}}\)

Theo ĐLBTKL :

\(m_{CaS}+m_{ddHBr}=m_{ddCaBr_2}+m_{H_2S}\)

=> \(2,16+50=m_{ddCaBr_2}+0,03\cdot34\)

=> \(m_{ddCaBr_2}=51,14\left(g\right)=m_2\)

=> \(C\%_{ddCaBr_2}=\frac{0,03\cdot200}{51,14}\cdot100\%\approx11,73\%\)

=> \(x\approx11,73\)

dễ nhưng vẫn nên check lại ...

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{Fe}=n_{H_2}=0,25\left(mol\right)\Rightarrow m_{Fe}=0,25.56=14\left(g\right)=m_1\)

\(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{18,25}{7,3\%}=250\left(g\right)=m_2\)

Đáp án A

Kết thúc phản ứng còn dư lại m1 (g) chất rắn Z

=> Z là Cu, dd Y gồm FeCl₂ và CuCl_2.

Fe₃O₄ + 8HCl → 2FeCl₃ + FeCl₂

0,01<---- 0,08       ------> 0,01

=> n C u = 0,02

2FeCl₃ + Cu → 2FeCl₂ + CuCl₂

0,02 -----> 0,01 --> 0,02

=> n C u dư = 0,01 => m1 = 6,4g

dd Y tác dụng với  A g N O 3

Ag⁺    +   Cl^-       → AgCl

0,08 <--- 0,08 ----> 0,08

Fe²⁺  +  Ag⁺  → Fe³⁺ + Ag

0,03     →                   0,03

=> m2 = 0,03.108 + 0,08.143,5 = 14,72g

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,04\left(mol\right)\Rightarrow m_1=m_{Zn}=0,04.65=2,6\left(g\right)\)

\(n_{HCl}=2n_{H_2}=0,08\left(mol\right)\Rightarrow m_{HCl}=0,08.36,5=2,92\left(g\right)\)

\(\Rightarrow m_2=m_{ddHCl}=\dfrac{2,92}{14,6\%}=20\left(g\right)\)

b, Ta có: m _{dd sau pư} = m_Zn + m _{dd HCl} - m_H2 = 22,52 (g)

\(n_{ZnCl_2}=n_{H_2}=0,04\left(mol\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,04.136}{22,52}.100\%\approx24,16\%\)

PTHH: 

\(M+H_2SO_4\rightarrow M_2\left(SO_4\right)_x+H_2\left(1\right)\)

\(M+O_2\rightarrow M_2O_x\left(2\right)\)

Phần 1:

\(n_{SO_4}=n_{H_2SO_4}=n_{H_2}=0,2\left(mol\right)\)

Ta có: \(m_{M_2\left(SO_4\right)_x}=m_M+m_{SO_4}\Leftrightarrow m_1=\dfrac{m}{2}+0,2.96=\dfrac{m}{2}+19,2\left(3\right)\)

Phần 2:

Ta có: \(m_O=m_{M_2O_x}-m_M=m_2-\dfrac{m}{2}\Rightarrow n_O=\dfrac{m_2}{16}-\dfrac{m}{32}\left(mol\right)\)

Lại có: \(n_{SO_4\left(1\right)}=x.n_{M_2\left(SO_4\right)_n}=\dfrac{x}{2}.n_M=x.n_{M_2O_x}=n_{O\left(2\right)}\)

\(\Leftrightarrow0,2=\dfrac{m_2}{16}-\dfrac{m}{32}\)

\(\Leftrightarrow3,2=m_2-\dfrac{m}{2}\)

\(\Leftrightarrow m_2=\dfrac{m}{2}+3,2\left(4\right)\)

Từ \(\left(3\right)\) và \(\left(4\right)\Rightarrow m_1-m_2=16\)

Đáp án B