\(m_{HCl}=219.10\%=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{ZnCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)

a, \(V_{H_2}=0,3.24,79=7,437\left(l\right)\)

b, \(m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

a, \(n_{H_2}=0,09\left(mol\right)\)

BT e, có: 2n_X = 2n_H2 ⇒ n_X = 0,09 (mol)

\(\Rightarrow\overline{M}_X=\dfrac{2,64}{0,09}=29,33\left(g/mol\right)\)

Mà: A, B thuộc 2 chu kì liên tiếp.

→ Mg và Ca.

Ta có: \(\left\{{}\begin{matrix}24n_{Mg}+40n_{Ca}=2,64\\2n_{Mg}+2n_{Ca}=0,09.2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,06\left(mol\right)\\n_{Ca}=0,03\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,06.24}{2,64}.100\%\approx54,55\%\\\%m_{Ca}\approx45,45\%\end{matrix}\right.\)

b, BTNT H, có: \(n_{H_2SO_4}=n_{H_2}=0,09\left(mol\right)\Rightarrow V_{ddH_2SO_4}=\dfrac{0,09}{2}=0,045\left(l\right)\)

BTNT Mg: n_MgSO4 = n_Mg = 0,06 (mol)

\(\Rightarrow C_{M_{MgSO_4}}=\dfrac{0,06}{0,045}=\dfrac{4}{3}\left(M\right)\)

\(S=\dfrac{40+m_{K_2CO_3\left(thêm\right)}}{150}\cdot100=30\\ \Rightarrow m_{K_2CO_3}=5g\\ \Rightarrow D\)

\(M+2HCl\rightarrow MCl_2+H_2\\ n_M=\dfrac{0,486}{M}mol\\ n_{H_2}=\dfrac{0,4958}{24,79}=0,02mol\\ \Rightarrow\dfrac{0,486}{M}=0,02\\ \Rightarrow M\approx24g/mol\\ \Rightarrow M.là.Mg\)

\(a,n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{8}{80}=0,1\left(mol\right)\\ b,n_{Fe_2\left(SO_4\right)_3}=\dfrac{m_{Fe_2\left(SO_4\right)_3}}{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{20}{400}=0,05\left(mol\right)\)

\(a.n_{CuO}=\dfrac{8}{80}=0,1mol\\ b.n_{Fe_2\left(SO_4\right)_3}=\dfrac{20}{400}=0,05mol\)

\(a.m_{H_2SO_4}=0,75.98=73,5g\)

\(b.m_{Cu}=0,6.64=38,4g\\ c.m_{\left(NH_4\right)_2SO_4}=0,8.132105,6g\)

CM(I2)=0.02M

gọi nồng độ iot chuyển hóa là x

  I2(k) <=> 2I(k)

pư: x 2x

cb: 0.02-x 2x

Ta có: (2x)^2/ 0.02-x = 3,80*10^-5

=> x=4,311*10^-4

=> [I2] = 0.02-4,311*10^-4=0,019M

=>[I] = 9,5*10^-3(M)