\(n_{Fe}=\dfrac{4.48}{56}=0.08\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.08....0.16.........0.08.....0.08\)

\(m_{HCl}=0.16\cdot36.5=5.84\left(g\right)\)

\(C\%_{HCl}=\dfrac{5.84}{800}\cdot100\%=0.73\%\)

\(m_{FeCl_2}=0.08\cdot127=10.16\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=4.48+800-0.08\cdot2=804.32\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{10.16}{804.32}\cdot100\%=1.26\%\)

\(HCl + AgNO_3 \to AgCl + HNO_3\\ n_{HCl} = n_{AgCl} = \dfrac{14,35}{143,5} = 0,1(mol)\\ \Rightarrow C\%_{HCl} = \dfrac{0,1.36,5}{10}.100\% = 36,5\%\)

Ta có: \(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

\(m_{H_2SO_4}=245.20\%=49\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)

PT: \(Fe_3O_4+4H_2SO_4\rightarrow FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,5}{4}\), ta được H₂SO₄ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe_2\left(SO_4\right)_3}=n_{Fe_3O_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(pư\right)}=4n_{Fe_3O_4}=0,4\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 23,2 + 245 = 268,2 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{0,1.152}{268,2}.100\%\approx5,67\%\\C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{268,2}.100\%\approx14,81\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1.98}{268,2}.100\%\approx3,65\%\end{matrix}\right.\)

Bạn tham khảo nhé!

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right);n_{H_2SO_4}=\dfrac{245.20\%}{98}=0,5\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2..........0,5

Lập tỉ lệ : \(\dfrac{0,2}{2}< \dfrac{0,5}{3}\)

=> H2SO4 dư

\(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

=> V H2 = 0,3.22,4= 6,72(l)

\(m_{ddsaupu}=5,4+245-0,3.2=249,8\left(g\right)\)

=> \(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{249,8}.100=13,69\%\)

a) mH2SO4=20%.245=49(g) ->nH2SO4=49/98=0,5(mol)

nAl=5,4/27=0,2(mol)

PTHH: 2Al +3 H2SO4 -> Al2(SO4)3 +3 H2

Ta có: 0,2/2 < 0,5/3

=> H2SO4 dư, Al hết, tính theo nAl

=> nH2SO4(p.ứ)=nH2=3/2. nAl=3/2. 0,2= 0,3(mol)

=> nH2SO4(dư)=0,5 - 0,3=0,2(mol)

=>mH2SO4(dư)=0,2.98=19,6(g)

b) V(H2,đktc)=0,3.22,4=6,72(l)

c) nAl2(SO4)3= 1/2. nAl=1/2. 0,2=0,1(mol)

=>mAl2(SO4)3=342.0,1=34,2(g)

mddAl2(SO4)3=mAl+ mddH2SO4-mH2=5,4+245 - 0,3.2= 249,8(g)

=>C%ddAl2(SO4)3= (34,2/249,8).100=13,691%

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)

c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)

a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\\ 2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)

Cho hỗn hợp tác dụng với NaOH, chất rắn không tan là Fe

=> m_Fe= 1,12 (g) \(\Rightarrow n_{Fe}=0,02\left(mol\right)\)

Ta có: \(n_{H_2\left(2\right)}=n_{Fe}=0,02\left(mol\right)\)

=> \(n_{H_2\left(1\right)}=\Sigma n_{H_2}-n_{H_2\left(2\right)}=0,065-0,02=0,045\left(mol\right)\)

\(\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2\left(1\right)}=0,03\left(mol\right)\)

\(\Rightarrow m_{Al}=0,03.27=0,81\left(g\right)\)

\(\Rightarrow\%m_{Al}=41,97\%,\%m_{Fe}=58,03\%\)

b) \(m_{FeCl_2}=0,02.127=2,54\left(g\right)\\ m_{AlCl_3}=0,03.133,5=4,005\left(g\right)\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)

nMgO=4,8/40=0,12(mol)

nHCl= (7,3%.300)/36,5= 0,6(mol)

PTHH: MgO + 2 HCl -> MgCl2 + H2O

Ta có: 0,6/2 > 0,12/1

-> MgO hết, HCl dư , tisnhh theo nMgO

=> Chất tan trong dd thu được có: MgCl2 và HCl(dư)

nMgCl2=nMgO=0,12(mol)

=> mMgCl2= 95.0,12=11,4(g)

nHCl(dư)= 0,6 -0,12.2=0,36(mol)

=>mHCl(dư)=0,36.36,5=13,14(g)

mddsau= mMgO + mddHCl= 4,8+300= 304,8(g)

=>C%ddMgCl2= (11,4/304,8).100=3,74%

C%ddHCl(dư)= (13,14/304,8).100=4,311%