a) PTPƯ nhiệt phân:
$KClO_3 \xrightarrow{t^o} KCl+\dfrac{3}{2}O_2$
$Ca(ClO_3)_2 \xrightarrow{t^o} CaCl_2+3O_2$
$Ca(ClO_2) \xrightarrow{t^o} CaCl_2+O_2$
$Ca(ClO_3)_2 → CaCl_2 + 3O_2$
b) PTHH:
$SO_2+\dfrac{1}{2}O_2 \xrightarrow{t^o;V_2O_5} SO_3$
$SO_3+H_2O \to H_2SO_4$
$⇒ n_{H_2SO_4}=n_{SO_2}=\dfrac{191,1.80\%}{98}=1,56$
$⇒ n_{SO_2}=\dfrac{1}{2}n_{H_2SO_4}=0,78$
Ta có:
$K_2CO_3+CaCl_2\to CaCO_3↓+2KCl$
BTKL:
$m_A-m_{O_2}=m_B=83,68 - 0,78.32=58,72$
$⇒ m_{K C l\ (B)} = m_B − m_{C a C l _2} = 38 , 74\ (g)$
$⇒ m_{KCl\ (D)} = 38,74 + 0,36.74,5 = 65,56\ (g)$
$m_{KCl\ (A)} = \dfrac{3}{22} .m_{KCl\ (D)}= 8,94\ (g)$
Khối lượng KCl trong $A$ là: $38,74 - 8,94 = 29,8\ (g)$
Theo pư $A ⇒ m_{KClO_3} = 49\ (g) ⇒ \%KClO_3 = 58,55\%$