Phản ứng xảy ra:
\(K_2O+H_2O\rightarrow2KOH\)
BTKL, \(m_{dd\left(spu\right)}=m_{K2O}+m_{H2O}=18,8+100=118,8\left(g\right)\)
\(n_{K2O}=\frac{18,8}{39.2+16}=0,2\left(mol\right)\)
\(\Rightarrow n_{KOH}=2n_{K2O}=0,4\left(mol\right)\)
\(\Rightarrow m_{KOH}=0,4.56=22,4\left(g\right)\)
\(C\%_{KOH}=\frac{22,4}{118,8}=18,86\%\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
\(n_{H2SO4}=\frac{1}{2}n_{KOH}=0,2\left(mol\right)\)
\(\Rightarrow m_{H2SO4}=0,2.98=19,6\left(g\right)\)
\(\Rightarrow m_{dd_{H2SO4}}=\frac{19,6}{10\%}=196\left(g\right)\)
\(n_{K_2O}=\frac{18,8}{94}=0,2\left(mol\right)\)
\(PTHH:K_2O+H_2O\rightarrow2KOH\)
(mol)______0,2____________0,4__
\(C\%_{KOH}=\frac{56.0,4}{18,8+100}.100\%=18,9\left(\%\right)\)
\(PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
(mol)_____0,4_______0,2___________________
\(m_{ddH_2SO_4}=\frac{0,2.98.100}{10}=196\left(g\right)\)
