\(n_{AlCl_3}=\dfrac{100\cdot5\%}{133.5}=\dfrac{10}{267}\left(mol\right)\)
\(n_{KOH}=\dfrac{400\cdot15\%}{56}=\dfrac{15}{14}\left(mol\right)\)
\(3KOH+AlCl_3\rightarrow3KCl+Al\left(OH\right)_3\)
\(\dfrac{10}{89}......\dfrac{10}{267}.......\dfrac{10}{89}......\dfrac{10}{267}\)
\(n_{KOH\left(dư\right)}=\dfrac{15}{14}-\dfrac{10}{89}=0.95\left(mol\right)\)
\(KOH+Al\left(OH\right)_3\rightarrow KAlO_2+2H_2O\)
\(\dfrac{10}{267}........\dfrac{10}{267}......\dfrac{10}{267}\)
\(n_{KOH\left(cl\right)}=0.95-\dfrac{10}{267}=0.92\left(mol\right)\)
\(m_{\text{dung dịch sau phản ứng}}=100+400=500\left(g\right)\)
\(C\%_{KOH}=\dfrac{0.92\cdot56}{500}\cdot100\%=10.304\%\)
\(C\%_{KCl}=\dfrac{\dfrac{10}{89}\cdot74.5}{500}\cdot100\%=1.67\%\)
\(C\%_{KAlO_2}=\dfrac{\dfrac{10}{267}\cdot98}{500}\cdot100\%=0.73\%\)
Ai đưa đề này cũng ác thiệt sự :((