\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
a) Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,1 0,3
\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
b) V = 100ml = 0,1l
\(C_{MH2SO4}=\dfrac{0,3}{0,1}=3\left(M\right)\)
\(m_{muối}=m_{Al2\left(SO4\right)3}=0,1.342=34,2\left(g\right)\)
Chúc bạn học tốt
a,2Al + 3H2SO4 → Al2(SO4)3 + 3H2
nAl = 5,4 : 27 = 0,2mol
nH\(_2\)=0,2.3:2 =0,3mol
VH\(_2\) = 0,3.22,4 =6,72 l
b. nH\(_2\)SO\(_4\) = 0,2.3:2=0,3mol
CM H\(_2\)SO\(_4\) = 0,3:0,1 =3M
nAl\(_2\)(SO\(_4\))\(_3\) = 0,2:2 =0,1mol
m\(Al_2 (SO_4 ) _3\) =0,1. 342 =34,2g
