Sửa đề: 1,4 (g) → 1,41 (g)
a, \(K_2O+H_2O\rightarrow2KOH\)
b, \(n_{K_2O}=\dfrac{1,4}{94}=0,015\left(mol\right)\)
Theo PT: \(n_{KOH}=2n_{K_2O}=0,03\left(mol\right)\Rightarrow C_{M_{KOH}}=\dfrac{0,03}{0,06}\left(M\right)\)
c, \(4K+O_2\rightarrow2K_2O\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{K_2O}=0,0075\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,0075.22,4=0,168\left(l\right)\)
\(A/K_2O+H_2O\rightarrow2KOH\\ B/n_{K_2O}=\dfrac{1,4}{94}mol\\ n_{KOH}=\dfrac{1,4}{94}\cdot2=\dfrac{7}{235}mol\\ C_{M_{KOH}}=\dfrac{7:235}{0,5}\approx0,06M\)
\(C/4K+O_2\xrightarrow[]{}2K_2O\\ n_{O_2}=\dfrac{1,4}{94}:2=\dfrac{7}{940}mol\\ V_{O_2,đktc}=\dfrac{7}{940}\cdot22,4\approx0,17l\\ V_{O_2,đkc}=\dfrac{7}{940}\cdot24,79\approx0,18l\)
