\(n_{Al_2\left(SO_4\right)_3}=\dfrac{200\cdot10\%}{342}=\dfrac{10}{171}\left(mol\right)\)
\(n_{NaOH}=\dfrac{500\cdot20\%}{40}=2.5\left(mol\right)\)
\(6NaOH+Al_2\left(SO_4\right)_3\rightarrow3Na_2SO_4+2Al\left(OH\right)_3\)
\(6..................1\)
\(2.5..................\dfrac{10}{171}\)
\(LTL:\dfrac{2.5}{6}>\dfrac{10}{171}\Rightarrow NaOHdư\)
\(n_{NaOH\left(dư\right)}=2.5-\dfrac{10}{171}\cdot6=2.15\left(mol\right)\)
\(NaOH+Al\left(OH\right)_3\rightarrow NaAlO_2+H_2O\)
\(\dfrac{20}{171}........\dfrac{20}{171}.........\dfrac{20}{171}\)
\(n_{NaOH\left(cl\right)}=2.15-\dfrac{20}{171}\approx2\left(mol\right)\)
\(m_{\text{dung dịch sau phản ứng}}=200+500=700\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{\dfrac{10}{57}\cdot142}{700}\cdot100\%=3.55\%\)
\(C\%_{NaAlO_2}=\dfrac{\dfrac{20}{171}\cdot82}{700}\cdot100\%=1.37\%\)
$n_{Al_2(SO_4)_3} = \dfrac{200.10\%}{342} = \dfrac{10}{171}(mol)$
$n_{NaOH} = \dfrac{500.20\%}{40} = 2,5(mol)$
$Al_2(SO_4)_3 + 6NaOH \to 2Al(OH)_3 + 3Na_2SO_4$
Ta thấy :
\(\dfrac{n_{Al_2(SO_4)_3}}{1} > \dfrac{n_{NaOH}}{6}\) nên NaOH dư
Theo PTHH :
n NaOH pư = 6n Al2(SO4)3 = 20/57(mol)
n Na2SO4 = 3n Al2(SO4)3 = 10/57(mol)
n Al(OH)3 = 2n Al2(SO4)3 = 20/171(mol)
Sau pư :
m dd = 200 + 500 - 78.20/171 = 709,122(gam)
\(C\%_{NaOH} = \dfrac{(2,5 - \dfrac{20}{57})40}{709,122}.100\% = 12,12\%\\ C\%_{Na_2SO_4} = \dfrac{\dfrac{10}{57}.142}{709,122}.100\% = 3,51\%\)
