\(n_{HCl}=\dfrac{100.36,5\%}{100\%}:36,5=1\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,02-->0,04----->0,02---->0,02

Xét: \(\dfrac{0,02}{1}< \dfrac{1}{2}\) => HCl dư

A gồm \(\left\{{}\begin{matrix}n_{HCl}=1-0,04=0,06\left(mol\right)\\n_{MgCl_2}=0,02\left(mol\right)\end{matrix}\right.\)

\(m_{dd.A}=0,02.24+100-0,02.2=100,44\left(g\right)\)

\(C\%_{HCl}=\dfrac{0,06.36,5.100\%}{100,44}=2,18\%\)

\(C\%_{MgCl_2}=\dfrac{0,02.95.100\%}{100,44}=1,89\%\)

\(m_{HCl}=\dfrac{100.36,5}{100}=36,5\left(g\right)\\ n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\\ n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\\ CaCO_3+2HCl\xrightarrow[]{}CaCl_2+CO_2+H_2O\\ \dfrac{0,1}{1}< \dfrac{1}{2}\Rightarrow HCl.dư\\ n_{CaCO_3}=n_{CaCl_2}=n_{CO_2}=0,1mol\\ m_{CaCl_2}=0,1.111=11,1\left(g\right)\\m_{CO_2}=0,1.44=4,4\left(g\right)\\ m_{ddCaCl_2}=10+100-4,4=105,6\left(g\right)\\ C_{\%CaCl_2}=\dfrac{11,1}{105,6}\cdot100\%\approx10,5\%\)

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

a. PTHH: \(Mg+2HCl--->MgCl_2+H_2\)

Theo PT: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)

=> \(m_{Mg}=0,2.24=4,8\left(g\right)\)

Theo PT: \(n_{HCl}=2.n_{Mg}=2.0,2=0,4\left(mol\right)\)

=> \(m_{HCl}=0,4.36.5=14,6\left(g\right)\)

=> \(C_{\%_{HCl}}=\dfrac{14,6}{200}.100\%=7,3\%\)

b. Ta có: \(m_{dd_{MgCl_2}}=4,8+200=204,8\left(g\right)\)

Theo PT: \(n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\)

=> \(m_{MgCl_2}=0,2.95=19\left(g\right)\)

=> \(C_{\%_{MgCl_2}}=\dfrac{19}{204,8}.100\%=9,28\%\)

Chọn D

Câu 3 : 

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

            1            2             1          1

          0,2         0,4          0,2         0,2

b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(m_{ddHCl}=\dfrac{14,6.100}{10,95}=133,3\left(g\right)\)

c) \(n_{H2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

d) \(n_{MgCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{MgCl2}=0,2.95=19\left(g\right)\)

\(m_{ddspu}=4,8+133,3-\left(0,2.2\right)=137,7\left(g\right)\)

\(C_{MgCl2}=\dfrac{19.100}{137,7}=13,8\)⁰/₀

 Chúc bạn học tốt

\(a.m_{HCl}=100.10\%+150.20\%=40\left(g\right)\\ C\%_{ddHCl}=\dfrac{40}{100+150}.100=16\%\\ b.n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{10,95\%.100}{36,5}=0,3\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,3}{1}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,3-2.0,1=0,1\left(mol\right)\\ m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\\ n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\\ m_{MgCl_2}=0,1.95=9,5\left(g\right)\\ m_{ddsau}=2,4+100-0,1.2=102,2\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{3,65}{102,2}.100\approx3,571\%\)

\(C\%_{ddMgCl_2}=\dfrac{9,5}{102,2}.100\approx9,295\%\)

phải là 2,24 lít khí `O_2` chứ bạn, nếu không thì rắn A tác dụng với HCl không tạo khí: )

a

\(2Mg+O_2\underrightarrow{t^o}2MgO\)

0,2<--0,1->0,2

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,1->0,2------>0,1-------->0,1

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

0,2--->0,2------>0,2

b

\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Xét: \(\dfrac{0,3}{2}>\dfrac{0,1}{1}\Rightarrow Mg.dư\)

\(n_{Mg.dư}=0,3-0,2=0,1\left(mol\right)\)

\(V_C=V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c

\(n_{HCl.đã.dùng}=\dfrac{100.29,2\%}{100\%}:36,5=0,8\left(mol\right)\)

\(n_{HCl.dư}=0,8-0,4=0,4\left(mol\right)\)

Dung dịch B gồm: \(\left\{{}\begin{matrix}n_{MgCl_2}=0,1+0,2=0,3\left(mol\right)\\n_{HCl.dư}=0,8-0,4=0,4\left(mol\right)\end{matrix}\right.\)

\(m_{ddB}=m_{Mg.dư}+m_{MgO}+m_{dd.HCl}-m_{H_2}=0,1.24+0,2.40+100-0,1.2=110,2\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{0,3.95.100\%}{110,2}=25,86\%\\ C\%_{HCl}=\dfrac{0,4.36,5.100\%}{110,2}=13,25\%\)