`a)PTHH:`

`Mg + H_2 SO_4 -> MgSO_4 + H_2`

`0,2`                                   `0,2`      `0,2`        `(mol)`

`n_[Mg]=[4,8]/24=0,2(mol)`

`b)m_[MgSO_4]=0,2.120=24(g)`

`c)C%_[MgSO_4]=24/[4,8+50-0,2.2].100~~44,12%`

100.\(\approx\)

Zn + 2HCl ---> ZnCl2 + H2 

0.2-→0.4------→0.2--→0.2 (mol)

nZn = 11,2\56 = 0.2(mol)

mZnCl2 = n*M = 0.2*127 = 25.4(g)

VH2(đktc) = n*22.4 = 0.2*22.4 = 4.48(l)

mHCl = n*M = 0.4*36.5 = 14.6(g)

C% =14,6\146*100% = 10(%)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{11,2}{65}=0,17\left(mol\right)\\ TheoPT:n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,17\left(mol\right)\\ m_{ZnCl_2}=0,17.136=23,12\left(g\right)\\ V_{H_2}=0,17.22,4=3,808\left(l\right)\\ c.n_{HCl}=2n_{Zn}=0,34\left(mol\right)\\ C\%_{HCl}=\dfrac{0,34.36,5}{146}.100=8,5\%\)

PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

Ta có: \(n_{KOH}=0,2\cdot0,5=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}=n_{K_2SO_4}=0,05\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{H_2SO_4}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\\C_{M_{K_2SO_4}}=\dfrac{0,05}{0,2+0,1}\approx0,17\left(M\right)\end{matrix}\right.\)

Bài 1 : 

200ml = 0,2l

100ml = 0,1l

\(n_{KOH}=0,5.0,2=0,1\left(mol\right)\)

a) Pt : \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O|\)

                2             1                1             2

              0,1          0,05            0,05

b) \(n_{H2SO4}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)

\(C_{M_{ddH2SO4}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)

c) \(n_{K2SO4}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

\(V_{ddspu}=0,2+0,1=0,3\left(l\right)\)

\(C_{M_{K2SO4}}=\dfrac{0,05}{0,3}=\dfrac{1}{6}\left(M\right)\)

 Chúc bạn học tốt

2NaOH + CuSO4 → Cu(OH)2 + Na2SO4

n NaOH = 0,2.5 = 1(mol)

n CuSO4 = 0,1.2 = 0,2(mol)

Ta có :

n NaOH / 2 = 0,5 > n CuSO4 / 1 = 0,2 => NaOH dư

n Cu(OH)2 = n CuSO4 = 0,2 mol

=> m A = 0,2.98 = 19,6 gam

n Na2SO4 = n CuSO4 = 0,2 mol

n NaOH pư = 2n CuSO4 = 0,4(mol)

V dd = 0,2 + 0,1 = 0,3(lít)

Suy ra:

CM Na2SO4 = 0,2/0,3 = 0,67M

CM NaOH = (1 - 0,4)/0,3 = 2M

\(n_{NaOH}=0.2\cdot0.5=0.1\left(mol\right)\)

\(n_{CuSO_4}=0.1\cdot2=0.2\left(mol\right)\)

\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\)

\(0.1.............0.05...............0.05...........0.05\)

\(m_{Cu\left(OH\right)_2}=0.05\cdot98=4.9\left(g\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.05}{0.2+0.1}=0.167\left(M\right)\)

\(C_{M_{CuSO_4\left(dư\right)}}=\dfrac{0.2-0.05}{0.1}=1.5\left(M\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

0,3     0,3             0,3           0,3 

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(m_{Fe}=0,3.56=16,8\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,3}{0,2}=1,5M\)

\(C_{M_{FeSO_4}}=\dfrac{n}{V}=\dfrac{0,3}{0,2}=1,5M\)

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

200ml = 0,2 lít.

a) Số mol H₂: n_H₂ = 6,72 ÷ 22,4 = 0,3 mol

Theo PTHH => Số mol Fe: n_Fe = 0,3 mol

     => Khối lượng Fe: m_Fe = 16,8g

b) Số mol H₂SO₄: n_H₂SO₄ = 0,3 mol

Nồng độ mol dd: C_M = 0,3 ÷ 0,2 = 1,5M

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

PTHH: Fe + H₂SO₄ ---> FeSO₄ + H₂

           0,3-->0,3----------->0,3

=> \(\left\{{}\begin{matrix}C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,15}=2M\\m_{muối}=0,3.152=45,6\left(g\right)\end{matrix}\right.\)

\(nHCl=0,2.0,3=0,06\\ 2Al+6HCl=>2AlCl3+3H2\\ =>nAl=0,02\left(mol\right)\\ =>mAl=0,02.27=0,54\left(g\right)\\ tacónAlCl3=0,02\left(mol\right)\\ =>Cm\left(AlCl3\right)=\dfrac{0,02}{0,2}=0,1\left(M\right)\)

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, \(n_{HCl}=0,2.0,3=0,06\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{1}{3}n_{HCl}=0,02\left(mol\right)\)

\(\Rightarrow m_{Al}=0,02.27=0,54\left(g\right)\)

c, \(n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,02\left(mol\right)\)

\(\Rightarrow C_{M_{AlCl_3}}=\dfrac{0,02}{0,2}=0,1\left(M\right)\)

\(a.PTHH:Fe_2O_3+6HCl--->2FeCl_3+3H_2O\)

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

Theo PT: \(n_{HCl}=6.n_{Fe_2O_3}=6.0,1=0,6\left(mol\right)\)

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

c. Theo PT: \(n_{FeCl_3}=2.n_{Fe_2O_3}=0,1.2=0,2\left(mol\right)\)

\(\Rightarrow m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)

Ta có: \(m_{dd_{FeCl_3}}=16+284=300\left(g\right)\)

\(\Rightarrow C_{\%_{FeCl_3}}=\dfrac{32,5}{300}.100\%=10,83\%\)

n_Fe2O3= 0.1(mol)

PTHH:  Fe2O3 + 6HCl -> 2FeCl3 + 3H2O (1)

a) Theo PT (1) : n_HCl = 6 n_Fe2O3 -> n_HCl = 0.1*6= 0.6(mol)

=> m_HCl= 0.6*36.5 = 21.9(g)

b)n_FeCl3=0.2(mol)

m_FeCl3= 162.5*0.2=32.5(g)

=> m_dd sau phản ứng: 248+16 = 264(g)

=> C%_muối= 32.5:264*100=12.3%

\(a.Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ b.n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\ n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\\ \Rightarrow m_{HCl}=21,9\left(g\right)\\ c.m_{ddsaupu}=16+284=300\left(g\right)\\ n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\\ C\%_{FeCl_3}=\dfrac{0,2.162,5}{300}.100=10,83\%\)