Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=0,1.0,5=0,05\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,05}{3}\), ta được Al dư.
Theo PT: \(n_{H_2}=n_{H_2SO_4}=0,05\left(mol\right)\) \(\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
\(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}=\dfrac{1}{60}\left(mol\right)\)
\(\Rightarrow C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{\dfrac{1}{60}}{0,1}=\dfrac{1}{6}\left(M\right)\)
theo phương trình : nAl2(SO4)3 \(=\dfrac{1}{3}.nH2SO4=\dfrac{1}{60}(mol)\)
CMAl2(SO4)3 \(=\dfrac{\dfrac{1}{60}}{0,1}=\dfrac{1}{6}(M)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) ; \(n_{H_2SO_4}=0,1.0,5=0,05\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
1/30 0,05 1/60 0,05
\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{1:60}{100}=\)0,0002M
