\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ 0,3.........0,3.........0,3.......0,3\left(mol\right)\\ m_{ddsau}=16,8+100=116,8\left(g\right)\\ m_{FeSO_4}=152.0,3=45,6\left(g\right)\\ C\%_{ddFeSO_4}=\dfrac{45,6}{116,8}.100\approx39,041\%\)

giup mk voi

\(nCuO=\dfrac{80}{80}=1\left(mol\right)\)

\(CuO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)

1             2                            1                        1

\(m_{\left(muối\right)}=1.182=182\left(g\right)\)

\(mCH_3COOH=2.60=120\left(g\right)\)

sao có 100g dd axit mà tới 120g CH3COOH ta

\(n_{Mg}=\dfrac{10,8}{24}=0,45\left(mol\right)\\ pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\) 
           0,45     0,45                          0,45 
\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\\ C\%_{H_2SO_4}=\dfrac{44,1}{176,4}.100\%=25\%\\ V_{H_2}=0,45.22,4=10,08\left(l\right)\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,4--->0,8------>0,4--->0,4

=> V_H2 = 0,4.22,4 = 8,96(l)

c) m_HCl = 0,8.36,5 = 29,2 (g)

=> \(m_{dd\left(HCl\right)}=\dfrac{29,2.100}{7,3}=400\left(g\right)\)

mdd (sau pư) = 22,4 + 400 - 0,4.2 = 421,6 (g)

=> \(C\%\left(FeCl_2\right)=\dfrac{127.0,4}{421,6}.100\%=12,05\%\)

\(n_{Al}=\dfrac{10,8}{27}=0,4(mol)\\ a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,6(mol);n_{Al_2(SO_4)_3}=0,2(mol)\\ b,V_{H_2}=0,6.22,4=13,44(l)\\ c,m_{dd_{H_2SO_4}}=\dfrac{0,6.98}{9,8\%}=600(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,2.342}{10,8+600-0,6.2}.100\%=11,22\%\)

a) \(n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(\text{mol}\right)\)

Phương trình hóa học phản ứng

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

_{2          :   3                   : 1               : 3} 

0,4        0,6            0,2             0,6

mol       mol           mol             mol

Thể tích khí H₂ sinh ra là 

\(V=n.22,4=0,6.22,4=13,44\left(l\right)\)

\(m_{Al_2\left(SO_4\right)_3}=n.M=0,2.342=68,4\left(g\right)\)

\(m_{H_2SO4}=n.M=0,6.98=58,8\left(g\right)\)

\(m_{H_2}=n.M=0,6.2=1,2\left(g\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{m_{H_2SO_4}.100\%}{C\%}=\dfrac{58,8.100\%}{9,8\%}=600\)(g)

=> \(m_{\text{dd sau pư}}=m_{ddH_2SO_4}+m_{Al}-m_{H_2}\)

= 600 + 10,8 - 1,2 (g) = 609,6 (g) 

=> \(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{m_{Al_2\left(SO_4\right)_3}}{m_{\text{dd sau pư}}}.100\%=\dfrac{68,4}{609,6}.100\%\)=11,22%

\(nCuO=\dfrac{80}{80}=1\left(mol\right)\)

\(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)

     2                     1                1                        1  (mol)

\(mCH_3COOH=2.60=120\left(g\right)\)

m muối = \(m\left(CH_3COO\right)_2Cu=1.182=182\left(g\right)\)

m H2O = 1.18 = 18 (g)

mdd = mddCH3COOH + m(CH3COO)2Cu + mH2O - mCuO

        = 100   + 182 + 18 - 80 = 220 (g)

\(C\%_{ddCH_3COOH}=\dfrac{120.100}{220}=54,55\%\)

a) \(n_{CuO}=\dfrac{80}{80}=1\left(mol\right)\)

PTHH: CuO + 2CH₃COOH ---> (CH₃COO)₂Cu + H₂O

              1---->2--------------------->1

=> m_muối = 1.182 = 182 (g)

b) \(C\%_{CH_3COOH}=\dfrac{60.2}{100}.100\%=120\%\) đề có sai không vậy bạn ?

a, \(n_{Fe}=\frac{0.56}{56}=0.01\left(mol\right)\)

 \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

  0.01      0.01       0.01       0.01

\(V_{H_2}=0.01\times22.4=0.224\left(l\right)\)

b, \(m_{H_2SO_4}=0.01\times98=0.98\left(g\right)\)

   \(m_{ddH_2SO_4}=\frac{100\times0.98}{19.6}=5\left(g\right)\)

\(m_{FeSO_4}=0.01\times152=1.52\left(g\right)\)

\(C\%_{FeSO_4}=\frac{1.52\times100}{5}=30.4\%\)

1)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

____0,1----->0,15

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%=\dfrac{14,7}{250}.100\%=5,88\%\)

2)

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

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

_______0,2------------------------------>0,2

=> V_CO2 = 0,2.22,4 = 4,48(l)

3)

\(n_A=\dfrac{18,4}{M_A}\left(mol\right)\)

PTHH: 2A + Cl₂ --t^o--> 2ACl

____\(\dfrac{18,4}{M_A}\)---------->\(\dfrac{18,4}{M_A}\)

=> \(\dfrac{18,4}{M_A}\left(M_A+35,5\right)=46,8=>M_A=23\left(Na\right)\)

4)

n_HCl = 0,2.3 = 0,6(mol)

PTHH: M + 2HCl --> MCl₂ + H₂

____0,3<-----0,6

=> \(M_M=\dfrac{7,2}{0,3}=24\left(Mg\right)\)