MgO+H2SO4->MgSO4+H2O

0,1-----0,1-------0,1--------0,1 mol

n MgO=4\40=0,1 mol

=>m H2SO4=0,1.98=9,8g

C%H2SO4=9,8\100 .100=9,8%

m MgSO4=0,1.120=12g

m muối =4+100-(0,1.18)=102,2g

=>C% muối=12\102,2.100=11,74 %

\(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.20\%}{98}=0,204\left(mol\right)\)

PTHH:

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

0,04      0,04           0,04    

\(\dfrac{0,04}{1}< \dfrac{0,204}{1}\) --> H2SO4 dư

\(C\%_{CuSO_4}=\dfrac{0,04.160}{3,2+100}.100\%=6,2\%\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2.98}{3,2+100}.100\%=19\%\)

B1 : nFe = 11,2 /56 = 0,2 (mol)

Fe+ 2HCl -- . FeCl2 + H2

mFeCl2 = 0,2.127 = 25,4 (g)

VH2 = 0,2 .22,4 = 4,48 (l)

mHCl = 0,4.36,5 = 14,6(g)

C%\(_{ddHCl}=\dfrac{ }{ }\)\(\dfrac{14,6.100}{280}=5,2\%\)

C2 :

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

nH2 = 17,92/22,4 = 0,8 (mol)

mAl = (2/3.0,8 ) .27 = 14,4 (g)

mAl2(SO4)3 = (1/3 . 0,8 ) . 342 = 91,2 (g)

mH2SO4 = 0,8 . 98 = 78,4 (g)

\(C\%_{ddH_2SO_4}=\dfrac{78,4.100}{120}=65,33\%\)

\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.1......0.3............0.1.........0.15\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

\(C\%_{HCl}=\dfrac{10.95}{100}\cdot100\%=10.95\%\)

\(m_{dd}=2.7+100-0.15\cdot2=102.4\left(g\right)\)

\(m_{AlCl_3}=0.1\cdot133.5=13.35\left(g\right)\)

\(C\%_{AlCl_3}=\dfrac{13.35}{102.4}\cdot100\%=13.04\%\)

\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

\(m_{H_2SO_4}=100.20\%=20\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{20}{98}=\dfrac{10}{49}\left(mol\right)\)

PTHH: CuO + H₂SO₄ → CuSO₄ + H₂O

Mol:      0,02      0,02          0,02

Ta có: \(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\) ⇒ CuO hết, H₂SO₄ dư

m_{dd sau pứ} = 1,6 + 100 = 101,6 (g)

\(C\%_{ddCuSO_4}=\dfrac{0,02.160.100\%}{101,6}=3,15\%\)

\(C\%_{ddH_2SO_4}=\dfrac{\left(\dfrac{10}{49}-0,02\right).98.100\%}{101,6}=17,76\%\)

a, \(n_{KOH}=\dfrac{44,8.25\%}{56}=0,2\left(mol\right);n_{H_2SO_4}=0,1.1,5=0,15\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ → K₂SO₄ + H₂O

Mol:       0,2         0,1              0,1

Ta có: \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\) ⇒ KOH hết, H₂SO₄ dư

 ⇒ Khi cho quỳ tím vào sẽ lm quỳ tím chuyển đỏ (vì trong dd vẫn còn axit)

b) \(m_{ddH_2SO_4}=1,1.100=110\left(g\right)\)

⇒ m_{dd sau pứ} = 44,8 + 110 = 154,8 (g)

\(C\%_{ddK_2SO_4}=\dfrac{0,1.174.100\%}{154,8}=11,24\%\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PTHH :

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

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

\(a,V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(b,m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(m_{ddHCl}=\dfrac{14,6.100}{10}=146\left(g\right)\)

\(c,m_{MgCl_2}=0,2.95=19\left(g\right)\)

\(m_{ddMgCl_2}=4,8+146-\left(0,2.2\right)=150,4\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{19}{150,4}.100\%\approx12,63\%\)

2. 

\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)

\(2K+2H_2O\rightarrow2KOH+H_2\uparrow\)

0,2                     0,2         0,1 

\(m_{KOH}=0,2.56=11,2\left(g\right)\)

\(m_{ddKOH}=7,8+100-\left(0,1.2\right)=107,6\left(g\right)\)

\(C\%=\dfrac{11,2}{107,6}.100\%\approx10,4\%\)

Cảm on ah

a) \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O

Mol:      0,4                         0,2        0,2

b) \(C\%_{ddHCl}=\dfrac{0,4.36,5.100\%}{100}=14,6\%\)

c) \(m_{CaCl_2}=0,2.101=20,2\left(g\right)\)

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