Câu 18:

$m_{HCl}=146\times 50\text{%}=73(g)$

$n_{HCl}=\dfrac{73}{36,5}=2(mol)$

$2HCl+Na_2CO_3→2NaCl+CO_2+H_2O$

  2                              →                           1

$V_{CO_2}=1\times 22,4=22,4(l)$

Câu 19:

$n_{NaOH}=0,5\times 9=4,5(mol)$

$3NaOH+FeCl_3→3NaCl+Fe(OH)_3$

  4,5                                                              1,5

$m_{Fe(OH)_3}=107\times 1,5=160,5(g)$

Câu 20:

$n_{Mg(OH)_2}=\dfrac{174}{58}=3(mol)$

$Mg(OH)_2\xrightarrow{t^o}MgO+H_2O$

 3                               3

$m_{MgO}=3\times 40=120(g)$

18)

Phản ứng xảy ra:

\(N{a_2}C{O_3} + 2HCl\xrightarrow{{}}2NaCl + C{O_2} + {H_2}O\)

Ta có:

\({m_{HCl}} = 146.50\%  = 73{\text{ gam}} \to {{\text{n}}_{HCl}} = \frac{{73}}{{36,5}} = 2{\text{ mol}}\)

\( \to {n_{C{O_{2}}}} = \frac{1}{2}{n_{HCl}} = 1{\text{ mol}} \to {{\text{V}}_{C{O_2}}} = 1.22,4 = 22,4{\text{ lít}}\)

Chọn \(C\)

19)

Phản ứng xảy ra:

\(F{\text{e}}C{l_3} + 3NaOH\xrightarrow{{}}F{\text{e}}{(OH)_3} + 3NaCl\)

Ta có:

\({n_{NaOH}} = 0,5.9 = 4,5{\text{ mol}} \to {{\text{n}}_{F{\text{e}}{{(OH)}_3}}} = \frac{1}{3}{n_{NaOH}} = 1,5{\text{ mol}}\)

\( \to {m_{F{\text{e}}{{(OH)}_3}}} = 1,5.(56 + 16.3 + 3) = 160,5{\text{ gam}}\)

Chọn \(C\)

20)

Phản ứng xảy ra:

\(Mg{(OH)_2}\xrightarrow{{{t^o}}}MgO + {H_2}O\)

Ta có:

\({n_{Mg{{(OH)}_2}}} = \frac{{174}}{{24 + 17.2}} = 3{\text{ mol = }}{{\text{n}}_{MgO}}\)

\( \to {m_{MgO}} = 3.(24 + 16) = 120{\text{ gam}}\)

Chọn \(A\)