Question

[Mai Anh Vũ]

Giúp mìk vs!!!

Answer 1

[Phạm Huỳnh Hà Anh]

Gọi 2 muối cacbonat của 2 kim loại kiềm thổ là \(MCO_3\).

\(MCO_3+2HCl\rightarrow MCl_2+H_2O+CO_2\uparrow\)

0.03 0.03 0.03(mol)\(m_{muối_{tăng}}=m_{MCl_2}-m_{MCO_3}=3.17-2.84=0.33\left(g\right)\)

\(M_{muối_{tăng}}=M_{Cl_2}-M_{CO_3}=71-60=11\left(\dfrac{g}{mol}\right)\)

\(n_{muối_{tăng}}=\dfrac{m}{M}=\dfrac{0.33}{11}=0.03\left(mol\right)\)

\(V_{CO_2}=n\times22.4=0.03\times22.4=0.672\left(l\right)\)

\(M_{MCO_3}=\dfrac{m}{n}=\dfrac{2.84}{0.03}\approx94.6\left(\dfrac{g}{mol}\right)\)

Mà \(M+CO_3=94.6\)

\(\Leftrightarrow M+60=94.6\)

\(\Rightarrow M=34.6\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow M_1\le34.6\le M_2\)

Mà M ở 2 chu kì liên tiếp.

\(\Rightarrow\left\{{}\begin{matrix}M_1:Mg\\M_2:Ca\end{matrix}\right.\)(chu kì 2, chu kì 3)

\(n_{HCl}=C_m\times V_{dd}\Rightarrow V_{dd}=\dfrac{n}{C_m}=\dfrac{0.06}{2}=0.03\left(l\right)\)

\(m_{ddHCl}=D\times V_{dd}=0.03\times1000\times1.05=31.5\left(g\right)\)