\(n_{hh}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(m_{Br_2}=200\cdot\dfrac{20}{100}=40\left(g\right)\)

\(n_{Br_2}=\dfrac{40}{160}=0.25\left(mol\right)\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(0.25........0.25..........0.25\)

\(\)\(n_{C_2H_4}=n_{hh}=0.25\left(mol\right)\)

=> Sai đề

PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Ta có: \(n_{C_2H_4}+n_{C_2H_2}=\dfrac{56}{22,4}=2,5\left(mol\right)\left(1\right)\)

Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=\dfrac{480}{160}=3\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_4}=2\left(mol\right)\\n_{C_2H_2}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{2.22,4}{56}.100\%=80\%\\\%V_{C_2H_2}=100-80=20\%\end{matrix}\right.\)

a) \(m_{tăng}=m_{C_2H_4}=0,84\left(g\right)\)

=> \(n_{C_2H_4}=\dfrac{0,84}{28}=0,03\left(mol\right)\)

Gọi số mol CH₄, H₂ trong 3360 ml A là a, b

=> \(a+b=\dfrac{3,36}{22,4}-0,03=0,12\left(mol\right)\) (1)

Gọi số mol CH₄, H₂ trong 0,7 lít hh là ak, bk

=> ak + bk + 0,03k = \(\dfrac{0,7}{22,4}=0,03125\) (2)

Và 16ak + 2bk + 0,84k = 0,4875 (3)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,09\left(mol\right)\\b=0,03\left(mol\right)\\k=\dfrac{5}{24}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,03}{0,15}.100\%=20\%\\\%V_{CH_4}=\dfrac{0,09}{0,15}.100\%=60\%\\\%V_{H_2}=\dfrac{0,03}{0,15}.100\%=20\%\end{matrix}\right.\)

b) 

\(n_{C_2H_4}=\dfrac{1,68}{22,4}.20\%=0,015\left(mol\right)\)

\(n_{CH_4}=\dfrac{1,68}{22,4}.60\%=0,045\left(mol\right)\)

\(n_{H_2}=\dfrac{1,68}{22,4}.20\%=0,015\left(mol\right)\)

Bảo toàn C: \(n_{CO_2}=0,075\left(mol\right)\)

Bảo toàn H: \(n_{H_2O}=0,135\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0,05.1=0,05\left(mol\right)\)

\(m_{ddCa\left(OH\right)_2}=1000.1,025=1025\left(g\right)\)

PTHH: Ca(OH)₂ + CO₂ --> CaCO₃ + H₂O

             0,05---->0,05----->0,05

            CaCO₃ + CO₂ + H₂O --> Ca(HCO₃)₂

           0,025<--0,025------------>0,025

\(m_{CaCO_3}=\left(0,05-0,025\right).100=2,5\left(g\right)\)

m_{dd sau pư} = 1025 + 0,075.44 + 0,135.18 - 2,5 = 1028,23 (g)

\(C\%_{Ca\left(HCO_3\right)_2}=\dfrac{0,025.162}{1028,23}.100\%=0,3939\%\)

$Zn + H_2SO_4 \to ZnSO_4 + H_2$

$n_{Zn} = n_{H_2} = \dfrac{1,12}{22,4} = 0,05(mol)$

$\%m_{Zn} = \dfrac{0,05.65}{5,25}.100\% = 61,9\%$

$\%m_{Cu} =1 00\% -61,9\% = 38,1\%$

Pthh:

\(Zn+H2SO4->ZnSO\text{4+H2}\)

\(nZn=nH2=\dfrac{1,12}{22,4}=0,05mol\)

\(=>mZn=0,05.65=3,25g\)\(=>\%mZn=\dfrac{3,25}{5,25}.100\%=62\%\)

\(=>\%mCu=100-62=38\%\)

PTHH: \(K_2O+2HCl\rightarrow2KCl+H_2O\)

            \(K_2SO_3+2HCl\rightarrow2KCl+SO_2\uparrow+H_2O\)

a) Ta có: \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{200\cdot14,6\%}{36,5}=0,8\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_3}=0,3\left(mol\right)\\n_{K_2O}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{K_2O}=\dfrac{0,1\cdot94}{0,1\cdot94+0,3\cdot158}\cdot100\%\approx16,55\%\\\%m_{K_2SO_3}=83,45\%\end{matrix}\right.\)

b) Theo các PTHH: \(n_{KCl}=0,8\left(mol\right)\) \(\Rightarrow m_{KCl}=74,5\cdot0,8=59,6\left(g\right)\)

Mặt khác: \(\left\{{}\begin{matrix}m_{hh}=56,8\left(g\right)\\m_{SO_2}=0,3\cdot64=19,2\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{SO_2}=237,6\left(g\right)\)

\(\Rightarrow C\%_{KCl}=\dfrac{59,6}{237,6}\cdot100\%\approx25,1\%\)

158 ở đâu ra vậy anh ?

\(m_Q=\left(7,7.2\right).0,1=1,54\left(g\right)\)

=> m_T = 1,54 (g)

Gọi \(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\\n_{H_2}=c\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a+b+c=0,11\left(1\right)\\16a+28b+2c=1,54\left(2\right)\end{matrix}\right.\)

n_giảm = n_H2(pư) = 0,11 - 0,1 = 0,01 (mol)

\(n_{Br_2}=\dfrac{4,8}{160}=0,03\left(mol\right)\)

Bảo toàn liên kết: b = 0,01 + 0,03 = 0,04 (mol) (3)

(1)(2)(3) => a = 0,02 (mol); b = 0,04 (mol); c = 0,05 (mol)

=> n_H2(Q) = 0,05 - 0,01 = 0,04 (mol)

=> \(\%V_{H_2}=\dfrac{0,04}{0,1}.100\%=40\%\)