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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

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

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

\(a)Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ b)Đặt:n_{Zn}=x\left(mol\right);n_{Fe}=y\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}65x+56y=12,1\\x+y=0,2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\\ \Rightarrow m_{Fe}=0,1.56=5,6\left(g\right);m_{Zn}=0,1.65=6,5\left(g\right)\\ c)n_{H_2SO_4}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow C\%_{H_2SO_{\text{ 4}}}=\dfrac{0,2.98}{196}.100=10\%\)