\(a.\)
\(n_{O_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
\(0.3...................................................0.15\)
\(m_{KMnO_4}=0.3\cdot158=47.4\left(g\right)\)
\(4R+nO_2\underrightarrow{t^0}2R_2O_n\)
\(\dfrac{0.6}{n}....0.15\)
\(M_R=\dfrac{19.5}{\dfrac{0.6}{n}}=32.5n\)
\(n=2\Rightarrow R=65\)
\(Rlà:Zn\)
\(a) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 2.\dfrac{3,36}{22,4} = 0,3(mol)\\ m_{KMnO_4} = 0,3.158 = 47,4(gam)\\ b) 4R + nO_2 \xrightarrow{t^o} 2R_2O_n\\ n_R = \dfrac{4}{n}n_{O_2} = \dfrac{0,6}{n}(mol)\\ \Rightarrow \dfrac{0,6}{n}R = 19,5\Rightarrow R = \dfrac{65}{2}n\)
Với n = 2 thì R = 65(Zn)
a) nO2=0,15(mol)
PTHH: 2 KMnO4 -to-> K2MnO4 + MnO2 + O2
0,3<---------------------------------------------------0,15(mol)
-> mKMnO4=0,3.158= 47,4(g)
ta có: \(n_{O_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
PTHH:
\(2KMnO_4\underrightarrow{t^o}MnO_2+O_2+K_2MnO_4\)
2mol 1mol
xmol 0.15mol
\(=>x=\dfrac{2\cdot0.15}{1}=0.3=n_{KMnO_4}\)
\(=>M_{KMnO_4}=0.3\cdot\left(39+55+16\cdot4\right)=47.4\left(g\right)\)
