Đáp án:

$\downarrow$

Giải thích các bước giải:

PTHH: $\text{$4P$ + $5O_2$ $\xrightarrow{t^o}$ $2P_2$$O_5$}$

a, Ta có: $\text{$n_P$ = $\dfrac{6,2}{31}$ = 0,2 mol}$

Theo PTHH $\rightarrow$ $\text{$n_{O_2}$ = $\dfrac{5}{4}$. $n_P$ = 0,25 mol}$

$\rightarrow$ $\text{$V_{O_{2(đktc)}}$ = 0,25. 22,4 = 5,6 lít}$

b, Do $\text{$V_{O_2}$ = $\dfrac{1}{5}$. $V_{kk}$}$

$\rightarrow$ $\text{$V_{kk_{(đktc)}}$ = $5$. $V_{O_{2(đktc)}}$ = 5,6. 5 = 28 lít}$

c, PTHH: $\text{$2KClO_3$ $\xrightarrow{t^o}$ $2KCl$ + $3O_2$}$

Ta thấy: $\text{$n_{KClO_3}$ = $\dfrac{2}{3}$. $n_{O_2}$ = $\dfrac{1}{6}$ mol}$

$\rightarrow$ $\text{$m_{KClO_3}$ = $\dfrac{1}{6}$. 122,5 = $\dfrac{245}{12}$ $\approx$ 20,41 gam}$

$\text{#Kudo}$

$\text{Đáp án+Giải thích các bước giải:}$

$\text{A.}$

$\text{4P + $5O_{2}$$\xrightarrow{t^o}$$2P_{2}O_{5}$}$

$\text{Ta có: $n_{P}$=$\dfrac{6,2}{31}$= 0,2 (mol)}$

$\text{Theo PT ta có: $n_{O_{2}}$=$\dfrac{5}{4}$$n_{P}$=$\dfrac{5}{4}$ × 0,2= 0,25 (mol)}$

$\text{⇒ $V_{O_{2}}$=0,25 × 22,4= 5,6 (lít)}$

$\text{B.}$

$\text{Vì oxi chiếm 20% thể tích không khí nên ta có:}$

$\text{$V_{KK}$=$\dfrac{5,6}{20}$ × 100= 28 (lít)}$

$\text{C.}$

$\text{$2KClO_{3}$$\xrightarrow{t^o}$2KCl + $3O_{2}$}$

$\text{Theo PT ta có: $n_{KClO_{3}}$=$\dfrac{2}{3}$$n_{O_{2}}$=$\dfrac{2}{3}$ × 0,25= $\dfrac{1}{6}$ (mol)}$

$\text{⇒ $m_{KClO_{3}}$=$\dfrac{1}{6}$ × 122,5≈ 20,42 (g)}$