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 $\to$ FeCl2 + H2

b) n Fe = 2,8/56 = 0,05(mol)

Theo PTHH :

n HCl = 2n Fe = 0,1(mol)

=> m dd HCl = 0,1.36,5/10% = 36,5(gam)

Ta có: \(n_{Fe}=\dfrac{2,24}{56}=0,04\left(mol\right)\)

a. PTHH: Fe + 2HCl ---> FeCl₂ + H₂

b. Theo PT: \(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,04\left(mol\right)\)

=> \(m_{FeCl_2}=0,04.127=5,08\left(g\right)\)

=> \(V_{H_2}=0,04.22,4=0,896\left(lít\right)\)

c. Theo PT: \(n_{HCl}=2.n_{Fe}=2.0,04=0,08\left(mol\right)\)

=> \(m_{HCl}=0,08.36,5=2,92\left(g\right)\)

Ta có: \(C_{\%_{HCl}}=\dfrac{2,92}{m_{dd_{HCl}}}.100\%=5\%\)

=> \(m_{dd_{HCl}}=58,4\left(g\right)\)

\(n_{Fe}=\dfrac{2,24}{56}=0,04mol\)

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

0,04    0,08      0,04       0,04

\(m_{FeCl_2}=0,04\cdot127=5,08\left(g\right)\)

\(V_{H_2}=0,04\cdot22,4=0,896\left(l\right)\)

\(m_{HCl}=0,08\cdot36,5=2,92\left(g\right)\)

\(m_{ddHCl}=\dfrac{2,92}{5}\cdot100=58,4\left(g\right)\)

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a) $CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
Theo PTHH : 

$n_{CO_2} = n_{CaCO_3} = \dfrac{10}{100} = 0,1(mol)$
$V_{CO_2} = 0,1.22,4 = 2,24(lít)$
b) $n_{HCl} = 2n_{CaCO_3} = 0,2(mol)$
$C_{M_{HCl}} = \dfrac{0,2}{0,25} = 0,8M$

c) $CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,1(mol)$
$m_{CaCO_3} = 0,1.100 = 10(gam)$

\(1,n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

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

           0,3---->0,6------------------>0,3

\(2,C_{M\left(HCl\right)}=\dfrac{0,6}{0,3}=2M\\ 3,V_{H_2}=0,3.22,4=6,72\left(l\right)\)

a,  Fe + 2HCl → FeCl₂ + H₂

b, ta có : n_Fe = 13.5/56 =0.24 mol =>> n_H2 =0.24 mol 

=>> V_H2 = 0.24x22.4= 5.376 (lít)

c, n_FeCl2 = n_{fe = 0.24} mol => m_FeCl2 = 0.24x 127= 30.48gam

lưu ý: t lấy số mol sắt là xấp xỉ do nó lẻ quá 

a) Fe + 2HCl -> FeCl2 + H2 (1)

b) nFe = 13,5 : 56 = 0,241 mol

Từ pt(1) => nH2 = nFe = 0,241 mol

Thể tích khí H2 là : V_H2=0,241 . 22,4 = 5,3984 l

c) Từ pt(1) => nFeCl2 = nFe = 0,241 mol

=> mFeCl2 = 0,241 . 217 = 30,607g

. 

a) Fe + 2HCl -> FeCl2  +  H2 ( 1 )

b) nFe = 13,5 : 56 = 0,241 mol

Từ pt(1) => nH2 = nFe = 0,241 mol

=> V_H2= 0,241 . 22,4 = 5,3984 l

c) Từ pt(1) => nFeCl2 = nH2 = 0,241 mol

=> mFeCl2 = 0,241 . 127 = 30,607g