\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

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

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

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{dd_{H_2SO_4}}=150\cdot1.08=162\left(g\right)\)

\(m_{dd}=6.5+162-0.1\cdot2=168.3\left(g\right)\)

\(C\%_{ZnSO_4}=\dfrac{0.1\cdot161}{168.3}\cdot100\%=9.56\%\)

a)

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{H_2SO_4} = n_{H_2} = n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)$

$V = 0,3.22,4 = 6,72(lít)$
$C_{M_{H_2SO_4}} = \dfrac{0,3}{0,25} = 1,2M$

b)

$n_{CuO} = \dfrac{16}{80} = 0,2(mol)$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$n_{CuO} < n_{H_2}$ nên $H_2$ dư

$n_{Cu} = n_{CuO} = 0,2(mol)$
$m_{Cu} = 0,2.64 = 12,8(gam)$

\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ PT:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

       0,3     0,3                        0,3 (mol)

a) V= n. 22,4 = 0,3 . 22,4 = 6,72(l)

 \(C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%=\dfrac{0,3.98}{200}.100\%=11,76\%\)

b) PT: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)

         0,3                0,3 

=> m_Cu=n.M=0,3.64=19,2(g)

n fe =0.02(mol)  

pthh: fe + h2so4 = feso4 + h2

(mol)0.02 0.02     ->0.02    0.02

=> m feso4 = 0.02 x 152= 3.04 (g)          m dd feso4 = (m fe + m dd h2so4) - m h2

                                                                                    = (1.12 + 200) -(2x 0.02)=201.08(g)

c% feso4 = 1.51%

 bạn nào giúp mình đi chớ giải cả bài mình không hiểu thế vào C% sau PƯ đâu

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

b, \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\)

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

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

a)

$n_{Al} = \dfrac{8,1}{27} = 0,3(mol)$

$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : 

$n_{H_2} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$

b) $n_{HCl} = 3n_{Al} = 0,9(mol)$
$\Rightarrow m_{dd\ HCl} = \dfrac{0,9.36,5}{3,65\%} = 900(gam)$

c)

$m_{dd\ sau\ pư}= 8,1 + 900 - 0,45.2 = 907,2(gam)$

$n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,15(mol)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{907,2}.100\% = 5,65\%$

a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

                  0,2--------->0,4--------------->0,4------->0,2

=> V_CO2 = 0,2.22,4 = 4,48 (l)

b) \(m_{dd.CH_3COOH}=\dfrac{0,4.60}{12\%}=200\left(g\right)\)

c) m_{dd sau pư} = 21,2 + 200 - 0,2.44 = 212,4 (g)

=> \(C\%_{muối}=\dfrac{0,4.82}{212,4}.100\%=15,44\%\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Ta có: \(n_{H_2SO_4}=0,2\cdot1,5=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2\cdot27=5,4\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\end{matrix}\right.\)