$$Cl_2+2KOH\xrightarrow[loãng ]{nguội}KCl+KClO+H_2O (1)$$$$3Cl_2+6KOH\xrightarrow[đậm đặc ]{nóng}5KCl+KClO_3+3H_2O (2)$$$$KClO_3\overset{MnO_2}{\rightarrow}KCl+\frac{3}{2}O_2\uparrow (3)$$a) $Cl^0-1e\rightarrow Cl^+$ $Cl^0+1e\rightarrow Cl^-$ $\Rightarrow Cl_2\rightarrow Cl^-+Cl^+$ $\Rightarrow Cl_2+2KOH\xrightarrow[loãng ]{nguội}KCl+KClO+H_2O$ $ 5.[Cl^0+1e\rightarrow Cl^-]$ $[Cl^0-5e\rightarrow Cl^{+5}]$ $3Cl_2\rightarrow 5Cl^-+Cl^{+5 }$ $\Rightarrow 3Cl_2+6KOH\xrightarrow[đậm đặc ]{nóng}5KCl+KClO_3+3H_2O$b)$n_{KCl(1)}=n_{KCl(2)}=x mol\Rightarrow n_{Cl_2(1)}=x mol;n_{Cl_2(2)}=\frac{3x}{5} mol$ $$\Rightarrow \frac{n_{Cl_2(1)}}{n_{Cl_2(2)}}=\frac{x}{\frac{3x}{5}}=\frac{5}{3}$$ c)$\frac{P_1V_1}{T_1}=\frac{P_0V_0}{T_0}\Rightarrow V_0=\frac{P_1V_1}{T_1}.\frac{T_0}{P_0}=\frac{70.11,94}{273+27}.\frac{273}{76}=10 lit$ $(2),(3)\Rightarrow n_{Cl_2}=2.n_{O_2}\Rightarrow V_{O_2}=\frac{V_{Cl_2}}{2}=\frac{10}{2}=5 lit$
$$Cl_2+2KOH\xrightarrow[loãng ]{nguội}KCl+KClO+H_2O (1)$$$$3Cl_2+6KOH\xrightarrow[đậm đặc ]{nóng}5KCl+KClO_3+3H_2O (2)$$$$KClO_3\overset{MnO_2}{\rightarrow}KCl+\frac{3}{2}O_2\uparrow (3)$$a) $Cl^0-1e\rightarrow Cl^+$ $Cl^0+1e\rightarrow Cl^-$ $\Rightarrow Cl_2\rightarrow Cl^-+Cl^+$ $\Rightarrow Cl_2+2KOH\xrightarrow[loãng ]{nguội}KCl+KClO+H_2O$ $ 5.[Cl^0+1e\rightarrow Cl^-]$ $[Cl^0-5e\rightarrow Cl^{+5}]$ $\Rightarrow 3Cl_2+6KOH\xrightarrow[đậm đặc ]{nóng}5KCl+KClO_3+3H_2O$b)$n_{KCl(1)}=n_{KCl(2)}=x mol\Rightarrow n_{Cl_2(1)}=x mol;n_{Cl_2(2)}=\frac{3x}{5} mol$ $$\Rightarrow \frac{n_{Cl_2(1)}}{n_{Cl_2(2)}}=\frac{x}{\frac{3x}{5}}=\frac{5}{3}$$ c)$\frac{P_1V_1}{T_1}=\frac{P_0V_0}{T_0}\Rightarrow V_0=\frac{P_1V_1}{T_1}.\frac{T_0}{P_0}=\frac{70.11,94}{273+27}.\frac{273}{76}=10 lit$ $(2),(3)\Rightarrow n_{Cl_2}=2.n_{O_2}\Rightarrow V_{O_2}=\frac{V_{Cl_2}}{2}=\frac{10}{2}=5 lit$
$$Cl_2+2KOH\xrightarrow[loãng ]{nguội}KCl+KClO+H_2O (1)$$$$3Cl_2+6KOH\xrightarrow[đậm đặc ]{nóng}5KCl+KClO_3+3H_2O (2)$$$$KClO_3\overset{MnO_2}{\rightarrow}KCl+\frac{3}{2}O_2\uparrow (3)$$a) $Cl^0-1e\rightarrow Cl^+$ $Cl^0+1e\rightarrow Cl^-$ $\Rightarrow Cl_2\rightarrow Cl^-+Cl^+$ $\Rightarrow Cl_2+2KOH\xrightarrow[loãng ]{nguội}KCl+KClO+H_2O$ $ 5.[Cl^0+1e\rightarrow Cl^-]$ $[Cl^0-5e\rightarrow Cl^{+5}]$
$3Cl_2\rightarrow 5Cl^-+Cl^{+5 }$ $\Rightarrow 3Cl_2+6KOH\xrightarrow[đậm đặc ]{nóng}5KCl+KClO_3+3H_2O$b)$n_{KCl(1)}=n_{KCl(2)}=x mol\Rightarrow n_{Cl_2(1)}=x mol;n_{Cl_2(2)}=\frac{3x}{5} mol$ $$\Rightarrow \frac{n_{Cl_2(1)}}{n_{Cl_2(2)}}=\frac{x}{\frac{3x}{5}}=\frac{5}{3}$$ c)$\frac{P_1V_1}{T_1}=\frac{P_0V_0}{T_0}\Rightarrow V_0=\frac{P_1V_1}{T_1}.\frac{T_0}{P_0}=\frac{70.11,94}{273+27}.\frac{273}{76}=10 lit$ $(2),(3)\Rightarrow n_{Cl_2}=2.n_{O_2}\Rightarrow V_{O_2}=\frac{V_{Cl_2}}{2}=\frac{10}{2}=5 lit$