$n_P=\dfrac{7,44}{31}=0,24mol \\PTHH : \\4P+5O_2\overset{t^o}\to 2P_2O_5(1) \\a.Theo\ pt\ (1) : \\n_{P_2O_5}=\dfrac{1}{2}.n_P=\dfrac{1}{2}.0,24=0,12mol \\⇒m_{P_2O_5}=0,12.142=17,04g \\b.Theo\ pt\ (1) : \\n_{O_2}=\dfrac{5}{4}.n_P=\dfrac{5}{4}.0,24=0,3mol \\⇒V_{O_2}=0,3.22,4=6,72l \\c.2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2(2) \\Theo\ pt\ (2) : \\n_{KMnO_4\ lt}=2.n_{O_2}=2.0,3=0,6mol \\Vì\ H=80\% \\⇒n_{KMnO_4\ tt}=\dfrac{0,6}{80\%}=0,75mol \\⇒m_{KMnO_4}=0,75.158=118,5g$