\(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)
PTHH: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
Mol (pt): 1 3 1 3
Mol (đề): 0,2
\(a,\) Theo pt, ta có: \(n_{H_2SO_4}=3\cdot n_{Al_2O_3}=3\cdot0,2=0,6\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{58,8}{200}\cdot100\%=29,4\%\)
\(b,\) Theo pt, ta có: \(n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,2\cdot342=68,4\left(g\right)\)
Theo ĐLBTKL: \(m_{\text{dd sau pứ}}=m_{Al_2O_3}+m_{\text{dd }H_2SO_4}=220,4\left(g\right)\)
\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{68,4}{220,4}\cdot100\%\approx31,03\%\)
\(\text{#}Toru\)
