Câu 1 : 

\(n_{Mg}=\dfrac{8.4}{24}=0.35\left(mol\right)\)

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

\(0.35.......0.7.........0.35..........0.35\)

\(C\%_{HCl}=\dfrac{0.7\cdot36.5}{146}\cdot100\%=17.5\%\)

\(m_{\text{dung dịch sau phản ứng}}=8.4+146-0.35\cdot2=153.7\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{0.35\cdot95}{153.7}\cdot100\%=21.6\%\)

Câu 2 :

\(n_{CaCO_3}=\dfrac{10}{100}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{114.1\cdot8\%}{36.5}=0.25\left(mol\right)\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(1................2\)

\(0.1.............0.25\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.25}{2}\Rightarrow HCldư\)

\(m_{\text{dung dịch sau phản ứng}}=10+114.1-0.1\cdot44=119.7\left(g\right)\)

\(C\%_{HCl\left(dư\right)}=\dfrac{\left(0.25-0.2\right)\cdot36.5}{119.7}\cdot100\%=1.52\%\)

\(C\%_{CaCl_2}=\dfrac{0.2\cdot111}{119.7}\cdot100\%=18.54\%\)

Câu 3 : 

\(n_{H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

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

\(0.3.......0.6......................0.3\)

\(m_{Fe}=0.3\cdot56=16.8\left(g\right)\)

\(m_{dd_{HCl}}=150\cdot1.2=180\left(g\right)\)

\(C\%_{HCl}=\dfrac{0.6\cdot36.5}{180}\cdot100\%=12.16\%\)

\(C_{M_{HCl}}=\dfrac{0.6}{0.15}=4\left(M\right)\)

\(n_{CaCO3}=\dfrac{10}{100}=0,1\left(mol\right)\)

Pt : \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O|\)

            1            2               1           1

          0,1         0,2              0,1

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

⇒ \(m_{HCl}=0,2.36,5=7,\left(g\right)\)

\(m_{dd}=\dfrac{7,3.100}{8}=91,25\left(g\right)\)

b) Nồng độ phần trăm của axit đề cho rồi đấy bạn , nên không cần tính nữa

c) \(n_{ZnCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{CaCl2}=0,1.111=11,1\left(g\right)\)

 Chúc bạn học tốt

a, nFe = 0,56/56 = 0,01 (mol)

PTHH: Fe + H2SO4 -> FeSO4 + H2

Mol: 0,01 ---> 0,01 ---> 0,01 ---> 0,01

mFeSO4 = 0,01 . 152 = 1,52 (g)

VH2 = 22,4 . 0,01 = 0,224 (l)

b, mH2SO4 = 0,01 . 98 = 0,98 (g)

c, mddH2SO4 = 0,98/19,6% = 5 (g)

d, mdd (sau p/ư) = 5 + 0,56 = 5,56 (g)

C%FeSO4 = 1,52/5,56 = 27,33%

a, nFe = 0,56/56 = 0,01 (mol)

PTHH: Fe + H2SO4 -> FeSO4 + H2

Mol: 0,01 ---> 0,01 ---> 0,01 ---> 0,01

mFeSO4 = 0,01 . 152 = 1,52 (g)

VH2 = 22,4 . 0,01 = 0,224 (l)

b, mH2SO4 = 0,01 . 98 = 0,98 (g)

c, mddH2SO4 = 0,98/19,6% = 5 (g)

d, mdd (sau p/ư) = 5 + 0,56 = 5,56 (g)

C%FeSO4 = 1,52/5,56 = 27,33%

n_CaCO3 = \(\dfrac{10}{100}=0,1\) mol

m_HCl = \(\dfrac{8\times114,1}{100}=9,128\left(g\right)\)

=> n_HCl = \(\dfrac{9,128}{36,5}=0,25\) mol

Pt: CaCO₃ + 2HCl --> CaCl₂ + H₂O + CO₂

...0,1 mol-> 0,2 mol-> 0,1 mol--------> 0,1 mol

Xét tỉ lệ mol giữa CaCO₃ và HCl:

\(\dfrac{0,1}{1}< \dfrac{0,25}{2}\)

Vậy HCl dư

m_HCl dư = (0,25 - 0,2) . 36,5 = 1,825 (g)

m_CO2 = 0,1 . 44 = 4,4 (g)

m_CaCl2 = 0,1 . 111 = 11,1 (g)

mdd sau pứ = m_CaCO3 + mdd HCl - m_CO2

....................= 10 + 114,1 - 4,4 = 119,7 (g)

C% dd HCl dư = \(\dfrac{1,825}{119,7}.100\%=1,52\%\)

C% dd CaCl₂ = \(\dfrac{11,1}{119,7}.100\%=9,27\%\)

\(n_{H_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

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

\(0.3.............................0.3..........0.45\)

\(m_{Al}=0.3\cdot27=8.1\left(g\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0.3\cdot342=102.6\left(g\right)\)

a,\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

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

Mol:      0,1            0,2                          0,1

\(m_{CaCO_3}=0,1.100=10\left(g\right)\)

b,\(C\%_{ddHCl}=\dfrac{0,2.36,5.100\%}{150}=4,87\%\)

c,m_{dd sau pứ}= 10+150-0,1.44 = 151,2 (g)

\(C\%_{ddCaCl_2}=\dfrac{0,1.111.100\%}{151,2}=7,34\%\)