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Ta có PTHH chung sau:

$\text{A2O3+3H2SO4$\rightarrow$A2(SO4)3+3H2O}$

Có $\text{nH2SO4=$\dfrac{29,4}{98}$=0,3 mol}$

Dựa vào PTHH) $\text{nA2O3=$\dfrac{0,3.2}{3}$=0,1 mol}$

$\rightarrow$ $\text{M A2O3=$\dfrac{16}{0,1}$=160}$

$\rightarrow$ $\text{M A2=160-48=112}$

$\rightarrow$ $\text{M A=$\dfrac{112}{2}$=56}$

$\text{a) Vậy A là Fe2O3}$

Có PTHH hoàn chỉnh sau) $\text{Fe2O3+3H2SO4$\rightarrow$Fe2(SO4)3+3H2O}$

Có $\text{nFe2(SO4)3=nFe2O3=0,1 mol}$

$\rightarrow$ $\text{b) mFe2(SO4)3=n.M=0,1.400=40g}$

$\text{Fe2O3+2CO$\xrightarrow{t^o}$2Fe+3CO2}$

Có $\text{nCO=2nFe2O3=0,1.2=0,2 mol}$

$\rightarrow$ $\text{V CO=n.22,4=0,2.22,4=4,48 lít}$

Đáp án:

\(\begin{array}{l}
a)A:Fe\\
b)\\
{m_{F{e_2}{{(S{O_4})}_3}}} = 40g\\
c)\\
{V_{CO}} = 6,72l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{A_2}{O_3} + 3{H_2}S{O_4} \to {A_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{29,4}}{{98}} = 0,3\,mol\\
{n_{{A_2}{O_3}}} = \dfrac{{0,3}}{3} = 0,1\,mol\\
{M_{{A_2}{O_3}}} = \dfrac{{16}}{{0,1}} = 160g/mol \Rightarrow 2{M_A} + 16 \times 3 = 160\\
 \Rightarrow {M_A} = 56g/mol \Rightarrow A:Fe\\
b)\\
{n_{F{e_2}{{(S{O_4})}_3}}} = {n_{F{e_2}{O_3}}} = 0,1\,mol\\
{m_{F{e_2}{{(S{O_4})}_3}}} = 0,1 \times 400 = 40g\\
c)\\
F{e_2}{O_3} + 3CO \xrightarrow{t^0} 2Fe + 3C{O_2}\\
{n_{CO}} = 3{n_{F{e_2}{O_3}}} = 0,3\,mol\\
{V_{CO}} = 0,3 \times 22,4 = 6,72l
\end{array}\)