\(a,CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ m_{rắn}=m_{MgO}=4g\\ m_{CaO}=9,6-4=5,6g\\ c,n._{CaO}=\dfrac{5,6}{56}=0,1mol\\ C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,1}{0,2}=0,5M\)

\(m_{ACl_3}=\dfrac{325.10}{100}=32,5g\\ n_{A_2O_3}=\dfrac{16}{2A+48}mol\\ n_{ACl_3}=\dfrac{32,5}{A+106,5}mol\\ A_2O_3+6HCl\rightarrow2ACl_3+3H_2O\\ \Rightarrow n_A=n_{ACl_3}:2\\ \Leftrightarrow\dfrac{16}{2A+48}=\dfrac{32,5}{A+106,5}:2\\ \Leftrightarrow A=56\)

Vậy A là Fe

\(a/n_{HCl}=0,3.2=0,6mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,2        0,6             0.2             0,3

\(V_{H_2}=0,2.22,4=4,48l\\ b/m_{Al}=0,2.27=5,4g\)

\(n_{H_2}=\dfrac{1,568}{22,4}=0,07mol\\ n_{Al}=a;n_{Mg}=b\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+24b=1,41\\1,5a+b=0,07\end{matrix}\right.\\ \Rightarrow a=0,03;b=0,025\\ m_{Al}=0,03.27=0,81g\\ m_{Mg}=1,41-0,81=0,6g\)

2)

\(n_{H_2}=\dfrac{14,874}{24,79}=0,6mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,4        1,2            0,4           0,6

\(a=m_{Al}=0,4.27=10,8g\\ V_{ddHCl}=\dfrac{1,2}{3}=0,4l\)

3)

\(a)n_{Na}=\dfrac{4,6}{23}=0,2mol\\ n_{O_2}=\dfrac{14,874}{24,79}=0,6mol\\ 4Na+2O_2\xrightarrow[]{t^0}2Na_2O\\ \Rightarrow\dfrac{0,2}{4}< \dfrac{0,6}{2}\Rightarrow O_2,dư\\ 4Na+2O_2\xrightarrow[]{t^0}2Na_2O\)

0,2          0,1         0,1

\(m_{O_2,dư}=\left(0,6-0,1\right).32=16g\\ b)m_{Na_2O}=0,1.62=6,2g\)

Gọi \(n_{Al,pư}=a\left(a>0\right)\)

\(Al+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3Ag\\ \Rightarrow m_{\uparrow}=108.3a-27a=28,5-27\\ \Rightarrow a=\dfrac{1}{198}mol\\ n_{Ag}=\dfrac{1}{198}\cdot3=\dfrac{1}{66}mol\\ m_{Ag}=\dfrac{1}{66}\cdot108=\dfrac{18}{11}=1,64g\)