a)
n K2O = 16,8/94 = 0,18 (mol)
PTHH: 4K + O2 ---> 2K2O (1)
0,36 0,09 0,18 (mol)
Theo PTHH(1), có:
n K=2nK2O =0,18.2 =0,36(mol)
=> mK=0,36.39=14,04(g)
b)
PTHH: 2KClO3 ---> 2KCl + 3O2 (2)
Theo Pthh(1) và (2) ,có:
nO2(2)=nO2(1)=0,09 (mol)
=> nKClO3 =2/3nO2(2)=2/3.0,09=0,06 (mol)
=> m KClO3=122,5. 0,06=7,35(g)
4K + O2 --to--➢ 2K2O (1)
a) \(n_{K_2O}=\dfrac{16,8}{94}=\dfrac{42}{235}\left(mol\right)\)
Theo PT1: \(n_K=2n_{K_2O}=2\times\dfrac{42}{235}=\dfrac{84}{235}\left(mol\right)\)
\(\Rightarrow m_K=\dfrac{84}{235}\times39=13,94\left(g\right)\)
b) 2KClO3 --to--➢ 2KCl + 3O2 (2)
Theo PT1: \(n_{O_2}=\dfrac{1}{2}n_{K_2O}=\dfrac{1}{2}\times\dfrac{42}{235}=\dfrac{21}{235}\left(mol\right)=n_{O_2\left(2\right)}\)
Theo PT2: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}\times\dfrac{21}{235}=\dfrac{14}{235}\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=\dfrac{14}{235}\times122,5=7,3\left(g\right)\)
em c.ơn trước ạ
