Đáp án:

 \(V_{dd}=350 ml\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(C{O_2} + C\xrightarrow{{{t^o}}}2CO\)

Ta có:

\({n_{C{O_2}}} = \frac{{5,6}}{{22,4}} = 0,25{\text{ mol}} \to {{\text{n}}_{CO}} = 2{n_{C{O_2}}} = 0,5{\text{ mol}}\)

Oxit có dạng \(RO\)

Ta có:

\({m_{O{\text{ trong oxit}}}} = 72.20\%  = 14,4{\text{ gam}} \to {{\text{n}}_O} = \frac{{14,4}}{{16}} = 0,9\;{\text{mol = }}{{\text{n}}_{RO}}\)

\( \to {m_R} = 72 - 14,4 = 57,6{\text{ gam}} \to {{\text{M}}_R} = \frac{{57,6}}{{0,9}} = 64 \to R:Cu\)

Ta có:

\({n_{CO}} = {n_{O{\text{ bi khử}}}} = 0,5{\text{ mol}}\)

Sau phản ứng rắn chứa \(Cu\) 0,9 mol và \(O\) 0,4 mol.

Bảo toàn e:

\(2{n_{Cu}} = 2{n_O} + 3{n_{NO}} \to 0,9.2 = 0,4.2 + 3{n_{NO}} \to {n_{NO}} = \frac{1}{3}\)

Ta có:

\({n_{Cu{{(N{O_3})}_2}}} = {n_{Cu}} = 0,9{\text{ mol}}\)

\( \to {n_{HN{O_3}}} = 2{n_{Cu{{(N{O_3})}_2}}} + {n_{NO}} = 0,9.2 + \frac{1}{3} = \frac{{32}}{{15}}{\text{ mol}}\)

\( \to {m_{HN{O_3}}} = \frac{{32}}{{15}}.63 = 134,4{\text{ gam}}\)

\( \to {m_{dd{\text{HN}}{{\text{O}}_3}}} = \frac{{134,4}}{{32\% }} = 420{\text{ gam}} \to {{\text{V}}_{dd}} = \frac{{420}}{{1,2}} = 350{\text{ ml}}\)

$n_{CO_2}=\dfrac{5,6}{22,4}=0,25(mol)$

$CO_2\buildrel{{+C}}\over\to 2CO$

$\Rightarrow n_{CO}=2n_{CO_2}=0,5(mol)$

Oxit kim loại: $RO$

$\%O=20\%\Rightarrow \dfrac{16.100}{R+16}=20$

$\Leftrightarrow R=64(Cu)$

Vậy oxit là $CuO$

$n_{CuO}=\dfrac{72}{80}=0,9(mol)$

$CuO\buildrel{{+CO}}\over\to Cu$

Sau phản ứng khử, $n_{Cu}=n_{CO}=0,5(mol)$

$n_{CuO\text{dư}}=0,9-0,5=0,4(mol)$

$3Cu+8HNO_3\to 3Cu(NO_3)_2+2NO+4H_2O$

$CuO+2HNO_3\to Cu(NO_3)_2+H_2O$

$\Rightarrow n_{HNO_3}=\dfrac{8}{3}n_{Cu}+2n_{CuO}=\dfrac{32}{15}(mol)$

$\Rightarrow m_{dd HNO_3}=\dfrac{32}{15}.63:32\%=420g$

$\to V_{dd}=\dfrac{420}{1,2}=350ml$