\(n_{KClO_3}=\dfrac{19,6}{122,5}=0,16mol\\ 2KClO_3\xrightarrow[]{t^0}2KCl+3O_2\\ n_{O_2\left(LT\right)}=\dfrac{0,16.3}{2}=0,24mol\\ m_{O_2\left(lt\right)}=0,24.32=7,68g\\ H=\dfrac{5,76}{7,68}\cdot100\%=75\%\)
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