\(m_{H2SO4}=14,7\left(g\right)\Rightarrow n_{H2SO4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,1 0,15 0,15
a) \(m_{Al}=0,1.27=2,7\left(g\right)\)
b) \(V_{H2\left(dkc\right)}=0,15.24,79=3,7185\left(l\right)\)
Chúc bạn học tốt
\(C\%_{ddH_2SO_4}=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{m_{ct}}{73,5}\cdot100\%=20\%\\ =>m_{H_2SO_4}=14,7\left(g\right)\\ n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{14,7}{98}=0,15\left(mol\right)\\ PTHH;2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\)
tỉ lệ 2 : 3 : 1 : 3
n(mol) 0,1<----0,15--------->0,05--------->0,15
`n_(Al)=n*M=0,1*27=2,7(g)`
\(V_{H_2\left(dkc\right)}=n\cdot22,4=0,15\cdot24,79=3,7185\left(l\right)\)
