Đáp án:
$a)\ V_{O_2}=4,48\ lít;\ m_{K_2MnO_4}=39,4\ g;\ m_{MnO_2}=17,4\ g;\ m_{O_2}=6,4\ g$
$b)\ m_P=4,96\ g$
Giải thích các bước giải:
PTHH: $2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 +O_2↑$
Ta có: $n_{KMnO_4}=\dfrac{63,2}{39+55+16.4}=0,4\ mol$
Theo PTHH: $n_{O_2}=n_{K_2MnO_4}=n_{MnO_2}=\dfrac{1}{2}.n_{KMnO_4}=0,2\ mol$
$→ V_{O_2}=0,2.22,4=4,48\ lít$
$m_{K_2MnO_4}=0,2.(39.2+55+16.4)=39,4\ gam$
$m_{MnO_2}=0,2.(55+16.2)=17,4\ gam$
$m_{O_2}=0,2.32=6,4\ gam$
PTHH: $4P +5O_2 → 2P_2O_5$
$→ n_P=\dfrac{4}{5}.n_{O_2}=0,16\ mol$
$→ m_P=0,16.31=4,96\ gam$