a, \(n_{Al_2O_3}=\dfrac{15,3}{102}=0,15\left(mol\right)\)
\(m_{H_2SO_4}=300.19,6\%=58,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)
PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,6}{3}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2SO_4\left(pư\right)}=3n_{Al_2O_3}=0,45\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,6-0,45=0,15\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,15.98=14,7\left(g\right)\)
b, Có lẽ đề hỏi khối lượng các chất sau pư chứ nhỉ?
Theo PT: \(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,15\left(mol\right)\\n_{H_2O}=3n_{Al_2O_3}=0,45\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)
\(m_{H_2O}=0,45.18=8,1\left(g\right)\)
mH2SO4 (dư) = 14,7 (g)
