Không ra thẳng được:

\(FeSO_4+2NaOH\to Fe(OH)_2\downarrow+Na_2SO_4\\ 4Fe(OH)_2+O_2+2H_2O\xrightarrow{t^o}4Fe(OH)_3\downarrow\)

\(1)n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ Fe+2HCl\to FeCl_2+H_2\)

Từ giả thiết và theo PT: 

\(\begin{cases} 24n_{Mg}+56n_{Fe}=5,2\\ n_{Mg}+n_{Fe}=0,15 \end{cases}\\ \Rightarrow n_{Mg}=0,1(mol);n_{Fe}=0,05(mol)\)

\(\Rightarrow \begin{cases} \%m_{Mg}=\dfrac{0,1.24}{5,2}.100\%=46,15\%\\ \%m_{Fe}=100-46,15=53,85\% \end{cases}\\ 2)\Sigma n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{1}=0,3(l)=300(ml)\)

lười làm thì đừng làm

box hóa có luật không tham khảo rồi

\(n_{Mg}=\dfrac{4,8}{24}=0,2(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ \Rightarrow n_{HCl}=0,4(mol)\\ 200ml=0,2l\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2(mol/l)=2(mol/dm^3)\\ \Rightarrow x=2\)

\(1)PTHH:CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow\\ n_{CaCO_3}=\dfrac{500.95\%}{100}=4,75(mol)\\ \Rightarrow n_{CaO}=4,75(mol)\\ \Rightarrow m_{CaO}=4,75.56=266(g)\\ \Rightarrow m_{CaO(tt)}=266.80\%=212,8(g)\\ m_{CaCO_3(k p/ứ)}=500.95\%.20\%=95(g)\\ \Rightarrow m_A=95+212,8=307,8(g)\\ 2)\%m_{CaO}=\dfrac{212,8}{307,8}.100\%=69,136\%\\ n_{CO_2}=n_{CaO}=4,75(mol)\\ \Rightarrow V_{CO_2}=4,75.22,4=106,4(l)\)

\(Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2\uparrow\\ n_{Fe}=\dfrac{1,12}{56}=0,02(mol)\\ \Rightarrow n_{CO}=0,03(mol)\\ \Rightarrow V_{CO}=0,03.22,4=0,672(l)=672(ml)\)

\(a,Zn+H_2SO_4\to ZnSO_4+H_2\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ \Rightarrow \begin{cases} 65n_{Zn}+27n_{Al}=3,68\\ n_{Zn}+1,5n_{Al}=0,1 \end{cases}\\\Rightarrow n_{Al}=n_{Zn}=0,04(mol)\\ \Rightarrow \%m_{Al}=\dfrac{0,04.27}{3,68}.100\%=29,345\&\\ \Rightarrow \%m_{Zn}=100-29,34=70,66\%\\ b,n_{H_2SO_4}=n_{H_2}=0,1\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,1.98}{10\%}=98(g)\)

\(c,n_{ZnSO_4}=0,04(mol);n_{Al_2(SO_4)_3}=0,02(mol)\\ \Rightarrow m_{ZnSO_4}=161.0,04=6,44(g)\\ m_{Al_2(SO_4)_3}=0,02.342=6,84(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{6,84}{3,68+98-0,1.2}.100\%=6,74\%\\ \Rightarrow C\%_{ZnSO_4}=\dfrac{6,44}{3,68+98-0,1.2}.100\%=6,35\%\)

\(A=3\cdot5^2\cdot\left(2\cdot3\cdot5\right)^n=3\cdot5^2\cdot2^n\cdot3^n\cdot5^n=2^n\cdot3^{n+1}\cdot5^{n+2}\)

Vì số ước của $A$ là $1030200$ nên:

\(\left(n+1\right)\left(n+1+1\right)\left(n+2+1\right)=1030200\\ \Rightarrow\left(n+1\right)\left(n+2\right)\left(n+3\right)=1030200\)

Lại có \(1030200=2^3\cdot3\cdot5^2\cdot17\cdot101=\left(2^2\cdot5^2\right)\cdot\left(3\cdot17\cdot2\right)\cdot101=100\cdot101\cdot102\)

Do đó \(\left\{{}\begin{matrix}n+1=100\\n+2=101\\n+3=102\end{matrix}\right.\Rightarrow n=99\)

Vậy \(A=75\cdot30^{99}\)

$x.(\dfrac{7}{4})^6=(\dfrac{7}{4})^8$

$\Rightarrow x=(\dfrac{7}{4})^8:(\dfrac{7}{4})^6=(\dfrac{7}{4})^{8-6}=(\dfrac{7}{4})^2=\dfrac{49}{16}$