Đáp án:
a) $m_{Fe}=3,36\ gam;\ m_{O_2}=1,28\ gam$
b) $m_{KClO_3}=3,267\ gam$
Giải thích các bước giải:
a/
PTHH: $3Fe +2O_2 \xrightarrow{t^o}Fe_3O_4$
Ta có: $n_{Fe_3O_4}=\dfrac{4,64}{56.3+16.4}=0,02\ mol$
Theo PTHH: $n_{Fe}=3n_{Fe_3O_4}=0,06\ mol → m_{Fe}=0,06.56=3,36\ gam$
$→m_{O_2}=m_{Fe_3O_4}-m_{Fe}=1,28\ gam$
Ta có: $n_{O_2}=\dfrac{1,28}{32}=0,04\ mol$
b/
$KClO_3\xrightarrow{t^o}KCl +\dfrac{3}{2}O_2↑$
$→ n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{0,·08}{3}$
$→ m_{KClO_3}=\dfrac{0,08}{3}.(39+35,5 +16.3)=3,267\ gam$