2KMnO4-to>K2MnO4+MnO2+O2
0,3--------------------------------------0,15 mol
4P+5O2-to>2P2O5
0,15--------0,06 mol
n KMnO4=\(\dfrac{47,4}{158}=0,3mol\)
=>VO2=0,15.22,4=3,36l
=>Pt2 : P dư=>m P dư
=>m P2O5= 0,06.142=8,52g
\(n_{KMnO_4}=\dfrac{47,4}{158}=0,3\left(mol\right)\\ 2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\\ n_{O_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ 1,\Rightarrow V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ 2,Ta.có:n_P=0,5\left(mol\right);n_{O_2}=0,15\left(mol\right)\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ Vì:\dfrac{0,5}{4}>\dfrac{0,15}{3}\Rightarrow Pdư\\ n_{P_2O_5}=\dfrac{2}{5}.n_{O_2}=\dfrac{2}{5}.0,15=0,06\left(mol\right)\\ \Rightarrow m_{sp}=m_{P_2O_5}=142.0,06=8,52\left(g\right)\)
