\(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\\
pthh:K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)
0,2 0,2 0,1
\(K_2O+H_2O\rightarrow2KOH\)
\(m_K=0,2.39=7,8\left(g\right)\\
m_{K_2O}=12,5-7,8=4,7\left(g\right)\\
\)
\(n_{K_2O}=\dfrac{4,7}{94}=0,05\left(mol\right)\\
n_{KOH}=2n_{K_2O}=0,1\left(mol\right)\\
\Sigma n_{KOH}=0,1+0,2=0,3\left(mol\right)\\
m_{KOH}=0,3.56=16,8\left(g\right)\)
2K+2H2O->2KOH+H2
0,2-------------0,2--------0,1
K2O+H2O->2KOH
0,05---------------0,1
n H2=0,1 mol
=>m K=0,2.39=7,8g
=>m K2O=4,7g=>n =0,05 mol
=>m KOH==0,3.56=16,8g
