Rất dễ em đừng lo lắng. Khi tăng áp suất cân bằng chuyển dịch theo chiều giảm số mol phân tử khí

\(n_{Na_2O}=\dfrac{a}{62}\left(mol\right)\)

\(m_{dd}=a+b\left(g\right)\)

`Na_2O + H_2O -> 2NaOH`

Theo PT: `n_{NaOH} = 2n_{Na_2O} = a/(31) (mol)`

\(\Rightarrow C\%_{NaOH}=\dfrac{40.\dfrac{a}{31}}{a+b}.100\%=10\%\)

\(\Leftrightarrow\dfrac{40a}{31}=0,1\left(a+b\right)\\ \Leftrightarrow\dfrac{369a}{310}=0,1b\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{0,1}{\dfrac{369}{310}}=\dfrac{31}{369}\)

Hay `a : b = 31 : 369`

\(Na_2O+H_2O->2NaOH\\ C\%=0,1=\dfrac{\dfrac{a}{62}\cdot2\cdot40}{a+b}\\ 0,1a+0,1b=\dfrac{40}{31}a\\ 0,1b=\dfrac{369}{310}a\\ a:b=\dfrac{0,1}{\dfrac{369}{310}}=\dfrac{31}{369}=0,08401\)

\(\left(NaCl,AlCl_3,FeCl_2,CuCl_2\right)-NaOH,dư->ddA\left(NaCl,NaAlO_2,NaOH\right),hhB\left(Fe\left(OH\right)_2,Cu\left(OH\right)_2\right)\\ hhB-t^0->Fe_2O_3,CuO-H_2,t^0->Fe,Cu-HCl,dư->\left(HCl,FeCl_2\right),Cu\\ Cu-Cl_2,dư->CuCl_2\\ FeCl_2\left(bay.hơi,cô.cạn\right)\\ ddA-CO_2,dư->Al\left(OH\right)_3,ddC\left(NaCl,NaHCO_3\right)\\ ddC-HCl,dư->NaCl\\ Al\left(OH\right)_3-HCl,dư->AlCl_3\left(bay.hơi,cô.cạn\right)\)

`n_{Cl} = 2n_{MgCl_2} = 2.0,525 = 1,05(mol)`

Số nguyên tử `Cl` là:

`1,05.6.10^{23} = 6,3.10^{23}` (nguyên tử)

\(n_{nguyên.tử.Cl}=2n_{MgCl_2}=2.0,525=1,05\left(mol\right)\)

`n_{FeSO_4} = 0,25(mol)`

\(n_{FeSO_4}=\dfrac{13,2.10^{23}}{6.10^{23}}=2,2\left(mol\right)\)

\(n_{NO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_A=n_{Al}+n_{Cu}=0,22+0,25=0,47\left(mol\right)\)

\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right);n_{N_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ \Rightarrow n_B=n_{O_2}+n_{N_2}=0,5+0,6=1,1\left(mol\right)\)

\(n_{Fe}=\dfrac{15.10^{23}}{6.10^{23}}=2,5\left(mol\right)\\ \Rightarrow n_C=n_{Cu}+n_{Fe}=0,25+2,5=2,75\left(mol\right)\\ n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right);n_{N_2}=\dfrac{2,7.10^{23}}{6.10^{23}}=0,45\left(mol\right)\\ \Rightarrow n_D=n_{O_2}+n_{CO_2}+n_{N_2}=0,25+0,5+0,45=1,2\left(mol\right)\)

\(1.\\ 2Al+6HCl->2AlCl_3+3H_2\\ Al_2O_3+6HCl->2AlCl_3+3H_2O\\ n_{Al}=\dfrac{2}{3}\cdot\dfrac{6,72}{22,4}=0,2mol\\ a.\%m_{Al}=\dfrac{0,2\cdot27}{36}=15\%\\ \%m_{Al_2O_3}=85\%\\ b.n_{Al_2O_3}=0,85\cdot\dfrac{36}{102}=0,3mol\\ C_M=\dfrac{0,6+1,8}{0,2}=12M\\ c.C_{M\left(AlCl_3\right)}=\dfrac{0,2+0,6}{0,2}=4\left(M\right)\)

\(2.\\ Zn+H_2SO_4->ZnSO_4+H_2\\ a.n_{Cu}=\dfrac{4,48}{22,4}=0,2mol\\ \%m_{Zn}=\dfrac{0,2\cdot65}{32,2}=40,37\%\\ \%m_{Cu}=59,63\%\\ b.C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,2}=1M\\ c.C_{M\left(ZnSO_4\right)}=\dfrac{0,2}{0,2}=1M\)

\(TQ:FeS_2->2SO_2->2SO_3->2H_2SO_4\\ m_{FeS_2}=\dfrac{1\cdot120}{98\cdot2\cdot0,7}=0,8746\left(\text{tấn }\right)\)

\(4-\dfrac{2\sqrt{3}}{2\sqrt{3}}=4-1=3\)