\(2Al+3H2SO4->Al2(SO4)3+3H2\)
\(a)nAl=\dfrac{5,4}{27}=0,2(mol)\)
\(nH2SO4=0,1.0,5=0,05(mol)\)
\(=>nH2=0,05(mol)\)
\(VH2=0,05.22,4=1,12(l)\)
\(b)nAl2(SO4)3=\dfrac{0,05}{3}=\dfrac{1}{60}(mol)\)
\(CMAl2(SO4)3=\dfrac{\dfrac{1}{60}}{0,1}=0,167(M)\)
\(100ml=0,1l\)
\(PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2mol\)
\(n_{H_2SO_4}=C_M.V_{dd}=0,5.0,1=0,05mol\)
Xét tỉ lệ số mol: \(\dfrac{0,2}{2}>\dfrac{0,05}{3}\Rightarrow\) \(Al\) phản ứng dư, \(H_2SO_4\) phản ứng hết
\(PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(n_{H_2}=n_{H_2SO_4}=0,05mol\)
a) \(V_{H_2}=n.22,4=0,05.22,4=1,12l\)
\(n_{Al_2\left(SO_4\right)_3}=\dfrac{0,05}{3}=\dfrac{1}{60}mol\)
b) \(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{\dfrac{1}{60}}{0,1}=0,17M\)
