a)

$n_{Zn} = \dfrac{13}{65} = 0,2(mol) ; n_{H_2 SO_4} = 0,5.2 = 1(mol)$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
Ta thấy : 

$n_{Zn} < n_{H_2SO_4}$ nên $H_2SO_4$ dư

$n_{ZnSO_4} = n_{H_2SO_4\ pư} = n_{Zn} = 0,2(mol)$
$m_{ZnSO_4} = 0,2.161=32,2(gam)$
$m_{H_2SO_4\ pư} = 0,2.98 = 19,6(gam)$

b)

$n_{H_2SO_4\ dư} = 1 - 0,2 = 0,8(mol)$
$C_{M_{H_2SO_4\ dư}} = \dfrac{0,8}{0,5} = 1,6M$
$C_{M_{FeSO_4}} = \dfrac{0,2}{0,5} = 0,4M$

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{H_2SO_4}=0,5.2=1\left(mol\right)\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ a.Vì:\dfrac{0,2}{1}< \dfrac{1}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{Zn}=0,2\left(mol\right)\\ m_{H_2SO_4\left(p.ứ\right)}=0,2.98=19,6\left(g\right)\\ m_{ZnSO_4}=161.0,2=32,2\left(g\right)\\ b.V_{ddsau}=V_{ddH_2SO_4}=0,5\left(l\right)\\ C_{MddZnSO_4}=\dfrac{0,2}{0,5}=0,4\left(M\right)\\ C_{MddH_2SO_4\left(dư\right)}=\dfrac{1-0,2}{0,5}=1,6\left(M\right)\)

a: \(n_{H_2SO_4}=0.25\cdot2=0.5\left(mol\right)\)

\(n_{Al_2O_3}=\dfrac{10.2}{27\cdot2+16\cdot3}=0.1\left(mol\right)\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

0,1              0,5

Vì 0,1/1<0,5/3

nên Al2O3 hết, H2SO4 dư

=>Tính theo Al2O3

b: 

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

0,1               0,3           0,1

\(m_{H_2SO_4\left(pư\right)}=0.3\cdot98=29.4\left(g\right)\)

\(m_{muối}=0.1\left(54+3\cdot96\right)=34.2\left(g\right)\)

Zn + 2HCl --> ZnCl2 + H2

0,5      1              0,5       0,5

nZn=\(\dfrac{32,5}{65}\)=0,5(mol)

mZnCl2=0,5.136=68(g)

VH2=0,5.22,4=11,2(l)

c) CM HCl=\(\dfrac{1}{0,25}=4M\)

nFe=3,92/56=0,07mol

mddCuSO4=200.1,12=224g

mCuSO4=224.10%=22,4g

nCuSO4=22,4/160=0,14mol

Fe+CuSO4-> Cu+FeSO4

0,07   0,14

TA CÓ 0,07/1 < 0,14/1 => CuSO4 dư

Fe +CuSO4-> Cu+FeSO4

0,07   0,07       0,07   0,07

mCu= 0,07.64=4,48g

b, C(FeSO4)=n/V= 0,07/0,2=0,35M

C(CuSO4) dư = n/V=0,07/0,2=0,35M

\(n_{Zn}=\dfrac{13}{65}=0,2mol\)

\(n_{H_2SO_4}=0,04.1=0,04mol\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

0,2  >   0,04                              ( mol )

0,04       0,04            0,04     0,04     ( mol )

\(m_{ZnSO_4}=0,04.161=6,44g\)

Câu b ko hiểu lắm bạn ơi!

\(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

\(n_{HCl}=2.0,2=0,4\left(mol\right)\)

PTHH :

                 \(CuO+2HCl\rightarrow CuCl_2+H_2\uparrow\)

trc p/ư:      0,15      0,4 

p/ư :          0,15      0,3       0,15          0,15 

sau p/ư :   0        0,1         0,15        0,15 

--> sau p/ư : HCl dư 

\(a,m_{CuCl_2}=0,15.135=20,25\left(g\right)\)

\(b,C_{M\left(CuCl_2\right)}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)

\(a)n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2\\ \dfrac{0,15}{1}< \dfrac{0,4}{2}\Rightarrow HCl.dư\\ n_{CuCl_2}=n_{CuO}=n_{H_2}=0,15mol\\ m_{CuCl_2}=0,15.135=20,25\left(g\right)\\ b)C_{MCuCl_2}=\dfrac{0,15}{0,2}=0,75\left(M\right)\\ n_{HCl\left(pư\right)}=0,15.2=0,3\left(mol\right)\\ n_{HCl\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\\ C_{MHCl\left(dư\right)}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

\(n_{HCl}=\dfrac{14,6\%.450}{36,5}=1,8\left(mol\right)\\ n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Vì:\dfrac{1,8}{6}>\dfrac{0,2}{1}\\ \Rightarrow HCldư\\ a.n_{FeCl_3}=0,2.2=0,4\left(mol\right)\\ m_{FeCl_3}=162,5.0,4=65\left(g\right)\\ b.n_{HCl\left(dư\right)}=1,8-6.0,2=0,6\left(mol\right)\\ m_{HCl\left(dư\right)}=0,6.36,5=21,9\left(g\right)\\ c.m_{ddsau}=32+450=482\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{21,9}{482}.100\approx4,544\%\\ C\%_{ddFeCl_3}=\dfrac{65}{482}.100\approx13,485\%\)

\(n_{Fe2O3}=\dfrac{32}{160}=0,2\left(mol\right)\)

\(m_{ct}=\dfrac{14,6.450}{100}=65,7\left(g\right)\)

\(n_{HCl}=\dfrac{65,7}{36,5}=1,8\left(mol\right)\)

Pt : \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O|\)

           1            6               2              3

         0,2          1,8            0,4

a) Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{1,8}{6}\)

                 ⇒ Fe₂O₃ phản ứng hết , HCl dư

                 ⇒ Tính toán dựa vào số mol của Fe₂O₃

\(n_{FeCl3}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

⇒ \(m_{FeCl3}=0,4.162,5=65\left(g\right)\)

b) \(n_{HCl\left(dư\right)}=1,8-\left(0,2.6\right)=0,6\left(mol\right)\)

⇒ \(m_{HCl\left(dư\right)}=0,6.36,5=21,9\left(g\right)\)

c) \(m_{ddspu}=32+450=482\left(g\right)\)

\(C_{FeCl3}=\dfrac{65.100}{482}=13,48\)⁰/₀

\(C_{HCl\left(dư\right)}=\dfrac{21,9.100}{482}=4,54\)⁰/₀

 Chúc bạn học tốt

a)\(n_{Fe_2O_3}=0,2\left(mol\right)\)

PT:\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

     \(0,2\)          \(1,2\)          \(0,4\)

\(\Rightarrow n_{FeCl_3}=0,4\left(mol\right)\)

\(\Rightarrow m_{FeCl_3}=65\left(g\right)\)

b) \(n_{HCl}=\dfrac{218.30\%}{35,5+1}=\dfrac{654}{365}\left(mol\right)\)

Từ PT \(\Rightarrow\)\(n_{HClpư}=1,2\left(mol\right)\)

\(\Rightarrow n_{HCldư}=\dfrac{654}{365}-1,2=\dfrac{216}{365}\left(mol\right)\)

\(\Rightarrow m_{HCldư}=21,6\left(g\right)\)

\(m_{dd}=32+218=250\left(g\right)\)

\(C\%_{FeCl_3}=\dfrac{65}{250}.100\%=26\left(\%\right)\)

\(C\%_{HCldu}=\dfrac{21,6}{250}.100\%=8,64\%\)

PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

a, Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\). ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

b, Ta có: \(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)

c, \(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5M\)

\(C_{M_{ZnSO_4}}=\dfrac{0,1}{0,2}=0,5M\)

Bạn tham khảo nhé!