a)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,1--->0,15------>0,05----->0,15
=> \(m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
=> \(m_{dd.H_2SO_4}=\dfrac{14,7.100}{9,8}=150\left(g\right)\)
b)
mdd sau pư = 2,7 + 150 - 0,15.2 = 152,4 (g)
\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
=> \(C\%=\dfrac{17,1}{152,4}.100\%=11,22\%\)
