a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, \(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,1.36,5}{10\%}=36,5\left(g\right)\)

c, \(n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05\left(mol\right)\)

Ta có: m _{dd sau pư} = 1,2 + 36,5 - 0,05.2 = 37,6 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,05.95}{37,6}.100\%\approx12,63\%\)

a) 

\(m_{H_2SO_4}=\dfrac{300.19,6}{100}=58,8\left(g\right)\)

=> \(m_{dd.H_2SO_4.9,8\%}=\dfrac{58,8.100}{9,8}=600\left(g\right)\)

=> \(m_{H_2O\left(thêm\right)}=600-300=300\left(g\right)\)

b)

\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

=> \(V_{dd.HCl.1,5M}=\dfrac{0,4}{1,5}=\dfrac{4}{15}\left(l\right)\)

=> \(V_{H_2O\left(thêm\right)}=\dfrac{4}{15}-0,2=\dfrac{1}{15}\left(l\right)=\dfrac{200}{3}\left(ml\right)\)

=> \(m_{H_2O\left(thêm\right)}=\dfrac{200}{3}.1=\dfrac{200}{3}\left(g\right)\)

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(0.2.......0.2......................0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(C\%H_2SO_4=\dfrac{0.2\cdot98\cdot100\%}{200}=9.8\%\)

\(n_{Mg}=\dfrac{2.4}{24}=0.1\left(mol\right)\)

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

\(0.1.......0.2...........0.1........0.1\)

\(m_{dd_{HCl}}=\dfrac{0.2\cdot36.5\cdot100}{14.6}=50\left(g\right)\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{MgCl_2}=0.1\cdot95=9.5\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=2.4+50-0.1\cdot2=52.2\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{9.5}{52.2}\cdot100\%=18.2\%\)

\(a.\\ C\%_{sau}=\dfrac{5}{100}=\dfrac{32.0,1}{32+m_{H_2O}}\\ m_{H_2O}=32\left(g\right)\\ b.\\ C_{M\left(sau\right)}=1=\dfrac{0,2.2}{0,2+V_{H_2O}}\\ V_{H_2O}=0,2\left(L\right)=200\left(mL\right)\)

\(n_{H_2}=\dfrac{0,896}{22,4}=0,04mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,04       0,08        0,04        0,04

\(m_{Zn}=0,04.65=2,6g\\ m_{ddHCl}=\dfrac{0,04.36,5}{14,6}\cdot100=10g\)

n_{Fe = 2,8 : 56 = 0,05 ( mol )} 

PTHH :      Fe +  2HCl -----> FeCl₂ + H₂

_mol             0,05    0,1                           0,05

^VH_{2 = 0,05 x 22,4 = 1,12 ( l )}

m_{HCl = 0,1 x 36,5 = 3,65 ( g)}

_=> m_{HCl (10%) = 3,65 x 100 : 10 = 36,5 (g)} 

PTHH 

      Fe      +    2HCl        \(\rightarrow\)         Fe  +    H₂O

gt 0,05           0,1                          0,05     0,05

m_Fe = 2,8 g \(\Rightarrow\)  n_Fe = \(\frac{m}{M}\)= \(\frac{2,8}{56}=0,05\left(mol\right)\)

Theo ptpư + gt ta có:

V\(H_2\) = n. 22,4 = 0,05 . 22,4 = 1,12 (lít)

m_HCl = 0,1 . 36,5 = 3, 65 (g)

m_{dd HCl} = \(\frac{3,65.100}{10}=36,5\left(g\right)\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

           0,05<-0,1<-----------0,05

=> m = 0,05.56 = 2,8 (g)

c) \(m_{HCl}=0,1.36,5=3,65\left(g\right)\Rightarrow m_{dd.HCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)

a) n Fe = 28/56 = 0,5(mol)

$Fe + 2HCl \to FeCl_2 + H_2$

Theo PTHH :

n HCl = 2n Fe = 1(mol)

=> m dd HCl = 1.36,5/10% = 365(gam)

b)

n FeCl2 = n H2 = n Fe = 0,5(mol)

Suy ra : 

V H2 = 0,5.22,4 = 11,2(lít)

m FeCl2 = 0,5.127 = 63,5(gam)

c)

Sau phản ứng: 

mdd = m Fe + mdd HCl - m H2 = 28 + 365 - 0,5.2 = 392(gam)

=> C% FeCl2 = 63,5/392   .100% = 16,2%

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