\(n_{K_2O}=\dfrac{m_1}{94}\left(mol\right)\)
PTHH: \(K_2O+2HCl\rightarrow2KCl+H_2O\)
\(\dfrac{m_1}{94}\)--->\(\dfrac{m_1}{47}\)------>\(\dfrac{m_1}{47}\)
=> \(\dfrac{m_1}{47}.74,5=m_1+1,65\)
=> m1 = 2,82 (g)
\(n_{HCl}=0,06\left(mol\right)\Rightarrow m_{HCl}=0,06.36,5=2,19\left(g\right)\)
=> \(m_2=\dfrac{2,19.100}{3,65}=60\left(g\right)\)
Ta có: \(m_{tăng}=m_1+1,65-m_1=1,65\left(g\right)\)
Ta có: \(n_{K_2O}=\dfrac{1,65}{71-16}=0,03\left(mol\right)\)
=> m1 = 0,03.94 = 2,82 (g)
PTHH: \(K_2O+2HCl\rightarrow2KCl+H_2O\)
0,03--->0,06
=> \(m_2=\dfrac{0,06.36,5}{3,65\%}=60\left(g\right)\)
