2Al + 3H2SO4 -----> Al2(SO4)3 + 3H2
nAl = 7,1/27 = 71/270 ( mol)
=> nH2 = 71/180 ( mol)
=> VH2= 8,86 lit
=> m muối=71\540 .342=44,967g
\(n_{Al}=\dfrac{7,1}{27}=\dfrac{71}{270}\left(mol\right)\\
pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(\dfrac{71}{270}\) \(\dfrac{71}{540}\) \(\dfrac{71}{180}\)
\(V_{H_2}=\dfrac{71}{540}.22,4=3l\\
m_{Al_2\left(SO_4\right)_3}=342.\dfrac{71}{180}=134,9g\)
a2Al+3H2SO4(dung dịch pha loãng)->Al2(SO4)3+3H2↑
b, nAl= 7,1/27=0,2 mol
Theo PTHH ta có: nH2=3/2 nAl= 3/2 . 0,2 = 0,3 mol
=> VH2(dktc)= 0,3 . 22,4= 6,72 lít
c, nAl2SO4= 1/2 . nAl= 1/2 . 0,2=0,1 mol
=> mAl2SO4=nAL2SO4+ MAl2SO4= 0,1 . 150 =15 g