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$3O_2\rightarrow 2O_3$ $O_3+2KI+H_2O\rightarrow 2KOH+I_2+O_2\uparrow$ $KOH+HCl\rightarrow KCl+H_2O$ $\frac{P_o.V_o}{T_o}=
\frac{P.V}{T} \Rightarrow V_o=\frac{2.4,928}{273+27,3}.\frac{273}{1}=8,96 lit\Rightarrow n_{O_2}=0,4 mol$ $\Rightarrow n_{O_3}=0,5n_{KOH}=0,5n_{HCl}=0,04 mol\Rightarrow n_{O_{2 pư}}=0,06 mol$ $\Rightarrow H=\frac{0,06.100}{0,4}=15\%$ $\Rightarrow n_{hh}=0,4-0,06+0,04=0,38 mol\Rightarrow p=2.\frac{0,38}{0,4}=1,9 atm$
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Trả lời 04-01-13 02:03 PM
viet130480
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