Đáp án + Giải thích các bước giải:

$a\big)$

$4R+O_2\xrightarrow{t^o}2R_2O$

Theo PT: $n_R=2n_{R_2O}$

$\to \dfrac{46,8}{M_R}=\dfrac{2.56,4}{2M_R+16}$

$\to M_R=39(g/mol)$

$\to R$ là kali $(K)$

$b\big)$

Bảo toàn KL:

$m_{O_2}=m_{K_2O}-m_K=56,4-46,8=9,6(g)$

$\to n_{O_2}=\dfrac{9,6}{32}=0,3(mol)$

$\to V_{kk}=\dfrac{0,3.22,4}{20\%}=33,6(l)$

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

$\text{a)}$

$4R + O_{2}\xrightarrow{t^o}2R_{2}O$

$\text{$n_{R}$=$\dfrac{m}{M}$=$\dfrac{46,8}{M_{R}}$ (mol)}$

$\text{$n_{R_{2}O}$=$\dfrac{m}{M}$=$\dfrac{56,4}{2M_{R} + 16}$ (mol)}$

$\text{Theo pt: $n_{R}$=2$n_{R_{2}O}$=2 × $\dfrac{56,4}{2M_{R} + 16}$= $\dfrac{112,8}{2M_{R} + 16}$}$

$\text{⇔ $\dfrac{46,8}{M_{R}}$ = $\dfrac{112,8}{2M_{R} + 16}$}$

$\text{⇔ 46,8(2$M_{R}$ + 16) = 112,8$M_{R}$}$

$\text{⇔ 93,6$M_{R}$ + 748,8 = 112,8$M_{R}$}$

$\text{⇔ 748,8 = 112,8$M_{R}$ - 93,6$M_{R}$}$

$\text{⇔ 19,2$M_{R}$ = 748,8}$

$\text{⇒ $M_{R}$=$\dfrac{748,8}{19,2}$= 39 (g/mol)}$

$\text{→ $M_{R}$ là $K$}$

$\text{b)}$

Theo định luật bảo toàn khối lượng:

$\text{$m_{K}$ + $m_{O_{2}}$ = $m_{K_{2}O}$}$

$\text{→$m_{O_{2}}$=$m_{K_{2}O}$ - $m_{K}$=56,4 - 46,8= 9,6 (g)}$

$\text{→ $n_{O_{2}}$=$\dfrac{m}{M}$=$\dfrac{9,6}{32}$= 0,3 (mol)}$

$\text{→ $V_{O_{2}(đktc)}$=0,3 × 22,4= 6,72}$ `(l)`

$\text{→ $V_{KK}$=6,72 × 5= 33,6}$ `(l)`