\(a)n_{CaCO_3^{}}^{} = \frac{{20}}{{100}} = 0,2\;{\text{mol}}\)

\(Ca(OH)_2^{} + CO_2^{} \to CaCO_3^{} + H_2^{}O\)

\({\text{                      0}}{\text{,2            0}}{\text{,2}}\)

\(d_{hh}^{} = 20,4.2 = 40,8g/mol\)

\(\frac{{CO}}{{CO_2^{}}} = \frac{{44 - 40,8}}{{40,8 - 28}} = \frac{1}{4}\)

\(n_{CO}^{} = \frac{1}{4}n_{CO_2^{}}^{} = \frac{{0,2}}{4} = 0,05\;{\text{mol}}\)

\( \to m_A^{} = 0,2.16 + 12 = 15,2g\)

\(72.n_{FeO}^{} + 160n_{Fe_2^{}O_3^{}}^{} = 15,2\)

\(n_{FeO}^{} + n_{Fe_2^{}O_3^{}}^{} = 0,15\)

\( \to n_{FeO}^{} = 0,1;n_{Fe_2^{}O_3^{}}^{} = 0,05\)

\(\% m_{FeO}^{} = \frac{{0,1.72}}{{15,2}}.100 = 47,36\% \)

\(\% m_{Fe_2^{}O_3^{}}^{} = 100 - 47,36 = 52,64\% \)

\(b){\text{cac phan ung di tu Fe}}{\text{,Fe}}_2^{}O_3^{};Fe_3^{}O_4^{};Fe_x^{}O_y^{}\)

\({\text{thi cho du bao nhieu Chat ve Fe khi qua HNO}}_3^{}\;{\text{du deu se tao Fe(NO}}_3^{})_3^{}\)

\( \to n_{Fe(NO)_3^{}}^{} = n_{Fe_{BanDau}^{}}^{} = (0,1 + 0,05.2).242 = 48,4g\)

Đáp án $+$ Giải thích các bước giải:

$a,$

Đặt $n_{FeO}=x(mol)\ ;\ n_{Fe_2O_3}=y(mol)$

$→x+y=0,15\ (1)$

$n_{CaCO_3}=\dfrac{20}{100}=0,2(mol)$

$\overline{M}_X=20,4×2=40,8(g/mol)$

$FeO+CO\xrightarrow{t^o}Fe+CO_2$

$Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2$

$B:\ Fe$ và $FeO,Fe_2O_3$ dư

$CO_2+Ca(OH)_2→CaCO_3+H_2O$

Theo PT: $n_{CO_2}=n_{CaCO_3}=0,2(mol)$

$\dfrac{n_{CO_2}}{n_{CO\ (dư)}}=\dfrac{40,8-28}{44-40,8}=4$

$⇒n_{CO\ (dư)}=\dfrac{0,2}{4}=0,05(mol)$

$n_{CO\ (pư)}=n_{CO_2}=0,2(mol)$

$n_{CO}=0,2+0,05=0,25(mol)$

Bảo toàn khối lượng:

$72x+160y+0,25×28=12+40,8×0,25\ (2)$

Từ $(1)(2)⇒x=0,1\ ;\ y=0,05$

$→\%m_{FeO}=\dfrac{0,1×72}{0,1×72+0,05×160}=47,36\%$

$→\%m_{Fe_2O_3}=100-47,36=52,64\%$

$b,$

Bảo toàn nguyên tố $Fe:\ n_{Fe(NO_3)_3}=0,1+0,05×2=0,2(mol)$

$→m_{muối}=0,2×242=48,4(g)$