Đáp án + Giải thích các bước giải:

Câu 1:

$n_{CO_2}=\frac{4,4}{44}=0,1(mol)$

$n_{O_2}=\frac{16}{32}=0,5(mol)$

$n_{H_2}=\frac{4}{2}=2(mol)$

$\to \overline{M}_X=\frac{4,4+16+4}{0,1+0,5+2}=\frac{122}{13}(g/mol)$

$\to d_{X/kk}=\dfrac{\frac{122}{13}}{29}\approx 0,32$

Câu 2:

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

$2KClO_3\xrightarrow{t^o,MnO_2}2KCl+3O_2$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

Giải thích các bước giải:

Bài 1:

$n_{CO_2}=\dfrac{4,4}{44}=0,1(mol)$

$n_{O_2}=\dfrac{16}{32}=0,5(mol)$

$n_{H_2}=\dfrac{4}{2}=2(mol)$

$\overline{M}=\dfrac{4,4+16+4}{0,1+0,5+2}=\dfrac{122}{13}(g/mol)$

$d_{(X)/kk)}=\dfrac{\dfrac{122}{13}}{29}=0,32$

Bài 2:

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

$2KClO_3\xrightarrow{t^o,\ MnO_2}2KCl+3O_2$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$