D = 1,1 g/ml mới đúng
\(n_{Al_2O_3}=\dfrac{1,02}{102}=0,01\left(mol\right)\)
\(m_{dd.H_2SO_4}=200.1,1=220\left(g\right)\)
\(n_{H_2SO_4}=\dfrac{220.4,9}{100}:98=0,11\left(mol\right)\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
0,01 ----> 0,03 ------> 0,01
Xét \(\dfrac{0,11}{1}< \dfrac{0,11}{3}\) => \(H_2SO_4\)dư
\(n_{H_2SO_4.dư}=0,11-0,03=0,08\left(mol\right)\Rightarrow CM_{H_2SO_4}=\dfrac{0,08}{0,2}=0,4M\)
\(n_{Al_2\left(SO_4\right)_3}=0,01\rightarrow CM_{Al_2\left(SO_4\right)_3}=\dfrac{0,01}{0,2}=0,05M\)
\(m_{dd.muối}=1,02+220=221,02\left(g\right)\)
\(C\%_{H_2SO_4}=\dfrac{0,08.98.100}{221,02}=3,55\%\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,01.100}{221,02}=1,55\%\)
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