Đáp án:

Giải thích các bước giải:

$a. Pt: 2KClO_3$ $\xrightarrow[MnO_2]{t^o}$ $2KCl + 3O_2(1)$

$n_{KClO_3}$= $\frac{24,5}{122,5}$= $0,2(mol)$

$\text{Từ pt(1):}$ $n_{O_2}$= $0,3(mol)$

$\rightarrow$ $V_{O_2}$= $0,3.22,4= 6,72(l)$

$b. Pt: 2O_2 + 3Fe$ $\xrightarrow{t^o}$ $Fe_3O_4(2)$

$\text{Từ pt(2):}$ $n_{Fe}$= $0,45(mol)$ ; $n_{Fe_3O_4}$= $0,15(mol)$

$\rightarrow$ $m_{Fe}$= $0,45.56= 25,2(g)$ ; $m_{Fe_3O_4}$= $0,15.232= 34,8(g)$

Đáp án + Giải thích các bước giải:

$a)$

$n_{KClO_3}=\frac{24,5}{122,5}=0,2(mol)$

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

$\to n_{O_2}=\frac{3}{2}n_{KClO_3}=0,3(mol)$

$\to V_{O_2}=0,3.22,4=6,72(l)$

$b)$

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

$\to n_{Fe}=\frac{3}{2}n_{O_2}=0,45(mol);n_{Fe_3O_4}=\frac{1}{2}n_{O_2}=0,15(mol)$

$\to m_{Fe}=0,45.56=25,2(g);m_{Fe_3O_4}=0,15.232=34,8(g)$