b,\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

PTHH: CuO + 2HCl → CuCl₂ + H₂

Mol:     0,2         0,4  

\(\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

c,\(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)

PTHH: ZnO + H₂SO₄ → ZnSO₄ + H₂

Mol:      0,2         0,2

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,2}{2}=0,1\left(l\right)\)

d,\(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)

PTHH: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Mol:     0,2           0,4

\(\Rightarrow V_{ddKOH}=\dfrac{0,4}{1}=0,4\left(l\right)\)

\(a,PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\\ \Rightarrow m_{CuSO_4}=0,1\cdot160=16\left(g\right)\\ b,n_{H_2SO_4}=n_{CuO}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{H_2SO_4}}=0,1\cdot98=9,8\left(g\right)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{9,8}{200}\cdot100\%=4,9\%\)

Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PT: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

a, \(n_{H_2SO_4}=3n_{Fe_2O_3}=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4}{9,8\%}=300\left(g\right)\)

b, \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 16 + 300 = 316 (g)

\(\Rightarrow C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{316}.100\%\approx12,66\%\)

Gọi\(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{FeS}=b\left(mol\right)\end{matrix}\right.\)

m_H2SO4 = 150.9,8% = 14,7 (g)

-> n_H2SO4 = \(\dfrac{14,7}{98}=0,15\left(mol\right)\)

\(n_{hhkhí\left(H_2,H_2S\right)}=\dfrac{\dfrac{224}{1000}}{22,4}=0,01\left(mol\right)\)

PTHH:

Fe + H₂SO₄ ---> FeSO₄ + H₂

a          a                 a           a

FeS + H₂SO₄ ---> FeSO₄ + H₂S

b             b                  b           b

Hệ phương trình\(\left\{{}\begin{matrix}56a+88b=0,72\\a+b=0,01\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,005\left(mol\right)\\b=0,005\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Fe}=0,005.56=0,28\left(g\right)\\m_{FeS}=0,005.88=0,44\left(g\right)\end{matrix}\right.\)

\(n_{H_2SO_4\left(pư\right)}=0,005+0,005=0,01\left(mol\right)\\ \Rightarrow n_{H_2SO_4\left(dư\right)}=0,15-0,01=0,14\left(mol\right)\\ m_{ddY}=0,72+150-0,005.2+0,005.34=150,88\left(g\right)\)

=> \(\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{152.\left(0,005+0,005\right)}{150,88}=1\%\\C\%_{H_2SO_4}=\dfrac{98.0,14}{150,88}=9,1\%\end{matrix}\right.\)

nCuO = 9,6/80 =0,12 ( mol )

nHCl= 150 . 6,327 / 100 /36,5 = 0,26 ( mol )

CuO + 2HCl --> CuCl2 + H2O
0,12 --> 0,24 --> 0,12

=> nHCl dư = 0,26 - 0,24 = 0,02 ( mol)

m dung dịch = 9,6 + 150 = 159,6 (g )

=> C% HCl dư= 0,02 . 36,5 /159,6 . 100 gần bằng 0,457 %

C% CuCl2 = 0,12 . 111/ 159,6 . 100 gần bằng 8,346 %

hì hì, tại mình nhớ nhầm qua cacl2 á

1: NaOH+HCl->NaCl+H2O

0,375        0,375

\(V_{HCl}=0.375\cdot22.4=8.4\left(lít\right)\)

\(C_{M\left(NaCl\right)}=\dfrac{0.375}{8.4+0.25}=\dfrac{15}{346}\)

\(n_{NaOH}=1,5.0,25=0,375\left(mol\right)\)

Pt : \(NaOH+HCl\rightarrow NaCl+H_2O\)

a) Theo Pt : \(n_{NaOH}=n_{HCl}=n_{NaCl}=0,375\left(mol\right)\)

\(V_{ddHCl}-\dfrac{0,375}{1,5}=0,25\left(l\right)\)

b) \(C_{MNaCl}=\dfrac{0,375}{0,25}=1,5\left(M\right)\)

 Chúc bạn học tốt 

\(a.n_{NaOH}=1,5.0,25=0,375mol\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{HCl}=n_{NaCl}=0,375mol\\ V_{HCl}=\dfrac{0,375}{1,5}=0,25l\\ b.C_{M\left(NaCl\right)}=\dfrac{0,375}{0,25+0,25}=0,75M\)

\(a,CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ \Rightarrow n_{H_2SO_4}=n_{CuSO_4}=n_{CuO}=0,2\left(mol\right)\\ b,m_{ddH_2SO_4}=\dfrac{0,2.98.100}{20}=98\left(g\right)\\ c,m_{ddCuSO_4}=16+98=114\left(g\right)\\ C\%_{ddCuSO_4}=\dfrac{0,2.160}{114}.100\approx28,07\%\)

a: n CuO=16/80=0,2(mol)

\(CuO+H_2SO_4\rightarrow CuSO_4\downarrow+H_2O\)

b: n CuO=0,2(mol)

=>n H2SO4=0,2(mol)

\(m_{ct\left(H_2SO_4\right)}=0.2\cdot\left(2+32+16\cdot4\right)=19.6\left(g\right)\)

=>m dd H2SO4=19,6/20%=98(g)

cho những bazơ sau :KOH, Ca(OH)2, Zn(OH)2, Cu(OH)2, Al(OH)3, Fe(OH)3.dảy các oxit bazơ nào sau đây tương ứng với các bazơ trên

A.K2O, Ca2O, ZnO, CuO, Al2O3,Fe3O4.

B.K2O, CaO, ZnO, Cu2O, Al2O3, Fe2O3.

C. K2O, CaO, ZnO, Cu2O, Al2O3, Fe2O3.

d. kết quả khác.

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