2KMnO4-->K2MnO4+MnO2+O2
a) n KMnO4=94,8/158=0,6(mol)
Theo pthh
n K2MnO4=1/2n KMnO4=0,3(mol)
m K2MnO4=0,3.197=59,1(g)
n MnO2=1/2n KMnO4=0,3(mol)
m MnO2=0,3.87=26,1(g)
m chất rắn = 26,1+59,1=85,2(g)
b) n O2=1/2n KMnO4=0,3(mol)
V O2=0,3.22,4=6,72(l)
Chúc bạn học tốt
a, PTHH : \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=\frac{m_{KMnO_4}}{M_{KMnO_4}}=\frac{94,8}{158}=0,6\left(mol\right)\)
- Theo PTHH : \(n_{K_2MnO_4}=n_{MnO_2}=\frac{1}{2}n_{KMnO_4}=\frac{1}{2}0,6=0,3\left(mol\right)\)
\(m_{MnO_2}=n_{MnO_2}.M_{MnO_2}=0,3\left(55+16.2\right)=0,3.87=26,1\left(g\right)\)
\(m_{K_2MnO_4}=n_{K_2MnO_4}.M_{K_2MnO_4}=0,3.\left(39.2+55+16.4\right)=59,1\left(g\right)\)
Ta có : mChất rắn = \(m_{MnO_2}+m_{K_2MnO_4}=26,1+59,1=85,2\left(g\right)\)
b, Theo PTHH : \(n_{O_2}=\frac{1}{2}n_{KMnO_4}=\frac{1}{2}0,6=0,3\left(mol\right)\)
-> \(V_{O_2}=n_{O_2}.22,4=0,3.22,4=6,72\left(l\right)\)
a) nKMnO4=\(\frac{94,8}{158}=0,6\left(mol\right)\)
PTHH:
2KMnO4\(\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,6→0,3→0,3→0,3 (mol)
=>mK2MnO4=0,3.197=59,1(g)
=>mMnO2=0,3.87=26,1(g)
=>mchất rắn=59,1+26,1=85,2(g)
b) VO2=0,3.22,4=6,72(l)
