a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\); \(n_{H_2SO_4}=\dfrac{200.24,5\%}{98}=0,5\left(mol\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,5}{3}\) => Al hết, H2SO4 dư
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2----->0,3---------->0,1------>0,3
=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b) mdd sau pư = 5,4 + 200 - 0,3.2 = 204,8 (g)
\(\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{204,8}.100\%=16,7\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,5-0,3\right).98}{204,8}.100\%=9,57\%\end{matrix}\right.\)
4al + 6h2so4--> 2al2(so4)3+3h2
mh2so4=24,5.200/100=49g
=>nh2so4=0,5
=>nAl=0,2
=> nH2=nAl=0,2
=>VH2=0,2.22,4=4,48
b)mdd=200+5,4-0,2.2=205
=>c%=mct/mdd.100=49/205.100=23,95%
