1)

a) 
PTHH: MgO + 2HCl --> MgCl₂ + H₂O

=> Số phân tử MgO : số phân tử HCl = 1 : 2

=> Số phân tử HCl = 2.3,6.10²³ = 7,2.10²³ (phân tử)

b) 

Số phân tử MgO : số phân tử MgCl₂ = 1 : 1

=> Số phân tử MgCl₂ = 3,6.10²³ (phân tử)

2)

\(m_{tăng}=m_{KL}-m_{H_2}\)

=> 1,54 = 1,68 - m_H2

=> m_H2 = 0,14 (g)

=> \(n_{H_2}=\dfrac{0,14}{2}=0,07\left(mol\right)\)

Gọi kim loại cần tìm là R

PTHH: R + 2HCl --> RCl₂ + H₂

          0,07<--------------0,07

=> \(M_R=\dfrac{1,68}{0,07}=24\left(g/mol\right)\)

=> R là Mg

đây là hóa lớp 8 mak bn

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

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

b, Ta có: \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,05\left(mol\right)\Rightarrow m_{MgCO_3}=0,05.84=4,2\left(g\right)\)

\(\Rightarrow m_{MgO}=5,2-4,2=1\left(g\right)\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  (1)

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

a) Ta có: \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)=n_{Mg}\) \(\Rightarrow m_{Mg}=1\cdot24=24\left(g\right)\)

\(\Rightarrow\%m_{Mg}=\dfrac{24}{32}\cdot100\%=75\%\) \(\Rightarrow\%m_{MgO}=25\%\)

b) Theo 2 PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{Mg}=2mol\\n_{HCl\left(2\right)}=2n_{MgO}=2\cdot\dfrac{32-24}{40}=0,4mol\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{HCl}=2,4mol\) \(\Rightarrow m_{ddHCl}=\dfrac{2,4\cdot36,5}{7,3\%}=1200\left(g\right)\)

c) Theo PTHH: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=1,2mol\)

\(\Rightarrow\Sigma m_{MgCl_2}=1,2\cdot95=114\left(g\right)\)

Mặt khác: \(m_{H_2}=1\cdot2=2\left(g\right)\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=1230\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{114}{1230}\cdot100\%\approx9,27\%\)

E xem lại đề 

\(a.PTHH:\)

\(Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)

\(MgO+2HCl--->MgCl_2+H_2O\left(2\right)\)

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

Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgO}=12-0,2.24=7,2\left(g\right)\)

\(\Rightarrow\%_{MgO}=\dfrac{7,2}{12}.100\%=60\%\)

c. Ta có: \(n_{hh}=0,2+\dfrac{7,2}{40}=0,38\left(mol\right)\)

Theo PT_(1,2): \(n_{HCl}=2.n_{hh}=2.0,38=0,76\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,76.36,5=27,74\left(g\right)\)

\(\Rightarrow m_{dd_{HCl}}=138,7\left(g\right)\)

\(\Rightarrow V_{dd_{HCl}}=126\left(ml\right)\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(Ca+2HCl\rightarrow CaCl_2+H_2\)

0,1                                  0,1     ( mol )

\(\rightarrow\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,1.40}{10}.100=40\%\\\%m_{MgO}=100\%-40\%=60\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{0,1.111}{10+390,2-0,1.2}.100=2,775\%\\C\%_{MgO}=\dfrac{4}{10+390,2-0,1.2}.100=1\%\end{matrix}\right.\)