Bài giải:

Dạng 4:

Bài 1:

a.$n_{H_2}=\frac{V_{H_2}}{22,4}= \frac{10,08}{22,4}=0,45(mol)$ 

$Fe+2HCl→FeCl_2+H_2↑$

  0,45        0,9          0,45    ←  0,45              (mol)

b. $m_{Fe}=n_{Fe}.M_{Fe}=0,45.56=25,2(g)$

c. $V_{dd..HCl}=150(ml)=0,15(l)$

$⇒CM_{HCl}=\frac{n_{HCl}}{V_{HCl}}=\frac{0,9}{0,15}=6(M)$     

Bài 2:

a. $n_{CO_2}=\frac{V_{CO_2}}{22,4}=\frac{6,72}{22,4}=0,3(mol)$ 

$CO_2+Ba(OH)_2→BaCO_3↓+H_2O$

  0,3              0,3              0,3                          (mol)

b. $V_{dd..Ba(OH)_2}=600(ml)=0,6(l)$

$CM_{Ba(OH)_2}=\frac{n_{Ba(OH)_2}}{V_{Ba(OH)_2}}=\frac{0,3}{0,6}=0,5(M)$    

c. $m_{kết..tủa}=m_{ BaCO_3}=n_{BaCO_3}.M_{BaCO_3}=0,3.233=69,9(g)$

 

1)

Phản ứng xảy ra:

\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)

Ta có:

\({n_{{H_2}}} = \frac{{10,08}}{{22,4}} = 0,45{\text{ mol = }}{{\text{n}}_{Fe}}\)

\( \to {m_{Fe}} = 0,45.56 = 25,2{\text{ gam}}\)

\( \to {n_{HCl}} = 2{n_{{H_2}}} = 0,45.2 = 0,9{\text{ mol}}\)

\( \to {C_{M{\text{ HCl}}}} = \frac{{0,9}}{{0,15}} = 6{\text{ lít}}\)

2)

Phản ứng xảy ra:

\(Ba{(OH)_2} + C{O_2}\xrightarrow{{}}BaC{O_3} + {H_2}O\)

Ta có:

\({n_{C{O_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol = }}{{\text{n}}_{Ba{{(OH)}_2}}} = {n_{BaC{O_3}}}\)

\( \to {C_{M{\text{ Ba(OH}}{{\text{)}}_2}}} = \frac{{0,3}}{{0,6}} = 0,5M\)

\( \to {m_{BaCO3}} = 0,3.233 = 69,9{\text{ gam}}\)