nAl=\(\dfrac{5,4}{27}\)=0,2 mol

2Al+3H₂SO₄→Al₂(SO₄)₃+3H₂

0,2-----0,3---------0,1-----------0,3

=>VH2=0,3.22,4=6,72l

=>CMH2SO4=\(\dfrac{0,3}{0,1}\)=3M

=>CM Al2(SO4)3=\(\dfrac{0,1}{0,1}\)=1M

nAl=\(\dfrac{5,4}{27}\)=0,2 mol

2Al+3H₂SO₄→Al₂(SO₄)₃+3H₂

0,2-----0,3---------0,1-----------0,3

=>VH2=0,3.22,4=6,72l

=>CMH2SO4=\(\dfrac{0,3}{0,1}\)=3M

=>CM Al2(SO4)3=\(\dfrac{0,1}{0,1}\)=1M

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

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

             0,2--->0,4--->0,2--->0,2

=> V = 0,2.22,4 = 4,48 (l)

b) \(C_{M\left(dd.HCl\right)}=\dfrac{0,4}{0,5}=0,8M\)

c) \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

1////          nAl=0,4mol   

2Al     +      6HCl -----> 2AlCl3 + 3H2

0,4mol      1,2mol         0,4mol    0,6 mol

a/ V_H2=0,6.22,4=13,44 l

b/ V=1,2/2=0,6 l

C_AlCl3=0,4/0,6=2/3 M

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

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,1                     0,1         0,1 

\(b,C_M=\dfrac{0,1}{0,1}=1M\)

\(c,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

Zn +H2SO4--->ZnSO4+H2(1)

Zn+2HCl----.ZnCl2 +H2(2)

a) Ta co

n\(_{Zn}\frac{3,25}{65}=0,05\left(mol\right)\)

Theo pthh 1,2

n\(_{H2}=n_{Zn}=0,05\left(mol\right)\)

V\(_{H2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)

b) Gọi n\(_{Zn}pt1=x\)

Gọi n\(_{Zn}pt2=y\)

=>x+y=0,05(*)

Mặt khác

m\(_{ZnSO4}=161x\)

m\(_{ZnCl2}=136y\)

=> 161x+136y=7,55(**)

Từ 1 và 2 ta có hệ pt

\(\left\{{}\begin{matrix}x+y=0,05\\161x+136y=7,55\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,03\\y=0,02\end{matrix}\right.\)

Theo pthh1

n\(_{H2SO4}=n_{Zn}=0,03\left(mol\right)\)

=>C\(_{M\left(H2SO4\right)}=\frac{0,03}{0,5}=0,06\left(M\right)\)

Theo pthh2

n\(_{HCl}=2n_{Zn}=0,02\left(mol\right)\)

C\(_{M\left(HCl\right)}=\frac{0,02}{0,5}=0,04\left(M\right)\)

Chúc bạn học tốt

Bài 1:

A . PTHH : 2AI + 3H₂SO₄ => AI₂(SO₄)₃ + 3H₂

B . nAI = 7.1 : 27 = 0.2 (mol)

 PT :2AI         +       3H₂SO₄      =>        AI₂(SO₄)₃       +   3H₂

        _{2 mol              3 mol                                                       3 mol}

        _{0.2 mol            0.3 mol                                                    0.3 mol}

=> V_{H2 = 22.4 X 0.3 = 6.72 (lít)}  

3.  500ml = 0.5 lít

Nồng độ mol/l của dd H₂SO₄ là:

C_{M H2SO4} = mol/lít =0.3/0.5 = 0.6 MOL/ LÍT

chuẩn nha hoàng kiều anh

a) 

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b)  \(n_{Zn=\frac{16,12}{65}=0,248\left(mol\right)}\)

Theo PTHH :  \(n_{H_2}=n_{Zn}=0,248\left(mol\right)\)

\(V_{H_2\left(đktc\right)}=0,248\times22,4=5,5552\left(l\right)\)

c ) Ta có :

Theo PTHH :  \(n_{HCl}=2n_{Zn}=0,496\left(mol\right)\)

\(\Rightarrow m_{HCl}=18,104\left(g\right)\)

\(m_{dd}=\frac{18,104\times100}{10}=181,04\left(g\right)\)

\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{2,84}{142}=0,02\left(mol\right)\)

PTHH : 

\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)

                0,04                     0,02                0,02 

\(a,C_M=\dfrac{n}{V}=\dfrac{0,04}{0,1}=0,4M\)

\(b,V_{H_2}=0,02.22,4=0,448\left(l\right)\)

\(c,PTHH:\)

\(CH_3COOH+C_2H_5OH\underrightarrow{t^o,H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)

0,04                                                         0,04 

\(m_{este}=0,04.90\%.88=3,168\left(g\right)\)

a) $Fe + 2HCl \to FeCl_2 + H_2$

 Theo PTHH : n H2 = n Fe = 8,4/56 = 0,15(mol)

V H2 = 0,15.22,4 = 3,36(lít)

b) n HCl = 2n Fe = 0,3(mol)

=> CM HCl = 0,3/0,2 = 1,5M

c) $CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Ta thấy :

n CuO = 32/80 = 0,4 > n H2 = 0,15 mol nên CuO dư

Theo PTHH : n Cu = n H2 = 0,15 mol

=> m Cu = 0,15.64 = 9,6 gam