a) 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

b) \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

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

____\(\dfrac{4}{15}\)<----0,4<--------------------0,4

=> \(m_{Al}=\dfrac{4}{15}.27=7,2\left(g\right)\)

c) \(C_{M\left(H_2SO_4\right)}=\dfrac{0,4}{0,15}=2,667M\)

a)

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
b)

$n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$n_{Al} = \dfrac{2}{3}n_{H_2} = 0,1(mol)$
$m_{Al} = 0,1.27 = 2,7(gam)$
c)

$n_{H_2SO_4} = n_{H_2} = 0,15(mol)$
$C\%_{H_2SO_4} = \dfrac{0,15.98}{100}.100\% = 14,7\%$

\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{3,36}{22,4}=0,15mol\)

a. PTHH: Fe + H₂SO₄ \(\rightarrow\) FeSO₄ + H₂

       TL:     1         1              1           1

      mol:  0,15  \(\leftarrow\) 0,15 \(\leftarrow\) 0,15  \(\leftarrow\) 0,15

\(b.m_{Fe}=n.M=0,15.56=8,4g\)

Đổi 150ml = 0,15 l

\(c.C_{MddH_2SO_4}=\dfrac{n}{V}=\dfrac{0,15}{0,15}=1M\)

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)

c, \(C\%_{H_2SO_4}=\dfrac{19,6}{50}.100\%=39,2\%\)

d, Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

Fe+2HCl->FeCl2+H2

0,3----0,6--0,3------0,3

n H2=0,3 mol

=>m Fe=0,3.56=16.8g

=>CM=0,6\0,1=6M

CuO+H2-tO>Cu+H2O

0.3---------------0,3

=>m Cu=0,3.64=19,2g

câu 2

FE+H2SO4->FeSO4+H2

0,25----0,25----0,25---0,25

n Fe=0,25 mol

=>mFeCl2=0,25.127=31,75g

=>VH2=0,25.22,4=5,6l

=>C% muối=\(\dfrac{31,75}{14+400-0,25.2}100=7,67\%\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

b) \(n_{Mg}=n_{H_2}=0,15\left(mol\right)\)

\(m_{Mg}=0,15.24=3,6\left(g\right)\)

c) \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)

\(C_{M_{HCl}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

0,2<---0,4<--------------0,2

a) \(m_{Zn.pứ}=0,2.65=13\left(g\right)\)

b) \(CM_{HCl.đã.dùng}=\dfrac{0,4}{0,1}=4M\)

a) Gọi số mol Al, Zn là a, b (mol)

=> 27a + 65b = 11,9 (1)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

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

             a----->1,5a----------------->1,5a

            Zn + H₂SO₄ --> ZnSO₄ + H₂

             b------>b------------------>b

=> 1,5a + b = 0,4 (2)

(1)(2) => a = 0,2 (mol); b = 0,1 (mol)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{11,9}.100\%=45,378\%\\\%m_{Zn}=\dfrac{0,1.65}{11,9}.100\%=54,622\%\end{matrix}\right.\)

b) n_H2SO4 = 1,5a + b = 0,4 (mol)

=> m_H2SO4 = 0,4.98 = 39,2 (g)

=> \(C\%_{dd.H_2SO_4}=\dfrac{39,2}{150}.100\%=26,133\%\)