\(n_{CuSO_4}=\dfrac{50}{250}=0.2\left(mol\right)\)

\(n_{FeSO_4}=\dfrac{27.8}{278}=0.1\left(mol\right)\)

\(C_{M_{CuSO_4}}=C_{M_{FeSO_4}}=\dfrac{0.1}{0.1964}=0.5\left(M\right)\)

\(m_{dd_A}=50+27.8+196.4=274.2\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{0.1\cdot160}{274.2}\cdot100\%=6.47\%\)

\(C\%_{FeSO_4}=\dfrac{0.1\cdot152}{274.2}\cdot100\%=5.54\%\)

\(n_{CuSO_4.5H_2O}=\dfrac{50}{250}=0,2\left(mol\right)\)

=> \(m_{CuSO_4}=0,2.160=32\left(g\right)\)

\(m_{H_2O}=0,2.5.18=18\left(g\right)\)

=> \(m_{FeSO_4}=0,1.152=15,2\left(g\right)\)

\(m_{H_2O}=0,1.7.18=12,6\left(g\right)\)

\(m_{dd}=196,4+50+27,8=274,2\left(g\right)\)

\(V_{dd}=\dfrac{196,4+18+12,6}{1000}=0,227\left(l\right)\)

=> \(CM_{CuSO_4}=\dfrac{0,2}{0,227}=0,72M\)

\(C\%_{CuSO_4}=\dfrac{32}{274,2}.100=11,67\%\)

\(CM_{FeSO_4}=\dfrac{0,1}{0,227}=0,44M\)

\(C\%_{CuSO_4}=\dfrac{15,2}{274,2}.100=5,54\%\)

Này không cho D hả em?

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(n_{NaOH}=2n_{Na_2O}=0,2\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

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

PTHH: 

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,15    0,3           0,15      0,15

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

\(a,n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

PTHH :

\(Na_2O+H_2O\rightarrow2NaOH\)

0,1           0,1        0,2

\(C_{M\left(A\right)}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

\(n_{K2O}=\dfrac{9,4}{94}=0,1\left(mol\right)\)

Pt : \(K_2O+H_2O\rightarrow2KOH|\)

         1           1              2

        0,1                        0,2

a) \(n_{KOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

200ml = 0,2l

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

b) Pt : \(KOH+HCl\rightarrow KCl+H_2O|\)

              1            1           1            1

            0,1          0,1

\(n_{HCl}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)

\(m_{ddHCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)

c) \(CO_2+2KOH\rightarrow K_2CO_3+H_2O|\)

        1          2               1              1

      0,05      0,1

\(n_{CO2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)

\(V_{CO2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)

 Chúc bạn học tốt

a)

$Zn + 2HCl \to ZnCl_2 + H_2$
$n_{ZnCl_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)$
$m_{ZnCl_2} = 0,1.136 = 13,6(gam)$
b)

$n_{HCl} = 2n_{Zn} = 0,2(mol) \Rightarrow C_{M_{HCl}} = \dfrac{0,2}{0,1} = 2M$

c)

CuO + H_2 \to Cu + H_2O$

$n_{CuO} = 0,125(mol) > n_{H_2} \to $ CuO$ dư

$n_{Cu} = n_{CuO\ pư} = n_{H_2} = 0,1(mol)$

$n_{CuO\ dư} = 0,125 - 0,1 = 0,025(mol)$

$\%m_{Cu} = \dfrac{0,1.64}{0,1.64 + 0,025.80}.100\% = 76,2\%$
$\%m_{CuO} = 23,8\%$

)

nHCl=2nZn=0,2(mol)⇒CMHCl=0,20,1=2MnHCl=2nZn=0,2(mol)⇒CMHCl=0,20,1=2M

c)

CuO + H_2 \to Cu + H_2O$

nCuO=0,125(mol)>nH2→nCuO=0,125(mol)>nH2→ CuO$ dư

nCu=nCuO pư=nH2=0,1(mol)nCu=nCuO pư=nH2=0,1(mol)

nCuO dư=0,125−0,1=0,025(mol)nCuO dư=0,125−0,1=0,025(mol)

a) Gọi số mol Zn, Al là a,b (mol)

=> 65a + 27b = 9,2 (1)

\(n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

             a---->2a-------------->a

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

            b----->3b--------------->1,5b

=> a + 1,5b = 0,25 (2)

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

=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{9,2}.100\%=70,65\%\\\%m_{Al}=\dfrac{0,1.27}{9,2}.100\%=29,35\%\end{matrix}\right.\)

b) n_HCl = 2a + 3b = 0,5 (mol)

=> \(C\%_{dd.HCl}=\dfrac{0,5.36,5}{200}.100\%=9,125\%\)

m_H2SO4= \(\dfrac{300.7,35}{100}=22,05g\)

n_H2SO4= \(\dfrac{22,05}{98}=0,225 mol\)

m_HCl= \(\dfrac{200.7,3}{100}=14,6g\)

n_HCl= \(\dfrac{14,6}{36,5}=0,4mol\)

                         H₂SO₄ + 2HCl → 2H₂O + Cl₂ ↑+ SO₂ ↑

n trước pư        0,225      0,4

n pư                  0,2  ←    0,4     →  0,4   → 0,2 → 0,2 mol

n sau  pư   dư 0,025       hết

a) m_Cl2= 0,2. 71= 14,2g

    m_SO2= 64. 0,2= 12,8g

    m_H2O= 18. 0,4=7,2g

m_{dd sau pư}=  300 +200 -14,2 -12,8= 473g

C%_{dd H2O}= \(\dfrac{7,2.100}{473}=1,52\)%

b) Mg + 2H₂O → Mg(OH)₂ + H₂ ↑

     x  →  2x     →      x        → x

     Fe + 2H₂O → Fe(OH)₂ + H₂↑

      y → 2y      →     y        → y

Gọi x,y lần lượt là số mol của Mg,Fe.

Ta có hệ phương trình:

    24x + 56y = 8,7              x= \(\dfrac{5}{64}\)

                                   ⇒

      2x   +  2y  = 0,4             y= \(\dfrac{39}{320}\)

V_H2= 22,4. \((\dfrac{5}{64}+\dfrac{39}{320})\)= 4,48l

m_{hh MG(OH)2, Fe(OH)2}= 8,7 +250 - 2.(\(\dfrac{5}{64}+\dfrac{39}{320}\)) = 2258,3g

m_Mg=24. \(\dfrac{5}{64}\)=1.875g

m_Fe= 8,7-1,875= 6,825g

\(n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ C_{M\left(HCl\right)}=\dfrac{0,1}{0,2}=0,5M\)

\(n_{Na}=\dfrac{6,9}{23}=0,3\left(mol\right)\\ m_{HCl}=200.3,65\%=7,3\left(g\right)\\ n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

\(PTHH:2Na+2HCl\rightarrow2NaCl+H_2\uparrow\\ LTL:0,3>0,2\Rightarrow Na.dư\)

Theo pt: n_H2 = 2n_HCl = 2.0,2 = 0,4 (mol)

V_H2 = 0,4.22,4 = 8,96 (l)

Theo pt: n_NaCl = n_{Na (phản ứng)} = n_HCl = 0,2 (mol)

=> \(\left\{{}\begin{matrix}m_{NaCl}=0,2.58,5=11,7\left(g\right)\\m_{Na\left(dư\right)}=\left(0,3-0,2\right).23=2,3\left(g\right)\\m_{H_2}=0,4.2=0,8\left(g\right)\end{matrix}\right.\)

=> \(m_{dd}=200+6,9-2,3-0,8=203,8\left(g\right)\)

=> C%_NaCl = \(\dfrac{11,7}{203,8}=5,74\%\)

2X + 2H2O => 2XOH + H2

nH2 = 0,015 mol => nX = 2nH2 = 0,03

=> M_X= 1,17/0,03 = 39 => Kali

2K+ 2H2O=> 2KOH + H2

KOH + HCl=> KCl + H2O

ta thấy nHCl=nKOH=n K = 0,03

=> C% HCl = \(\frac{0,03.36,5}{200}\) . 100% = 0,5475%

PTHH: 2X+2H₂O ---->2XOH +H₂

_{mol: 0.03 <--0.03 0.015}

_{---->>X=1.17/0.03=39(K)}

_{KOH + HCl ------> KCl + H2}

_{mol: 0,03--->0,03}

C%HCl= 0,03.36,5:200=0,547%