pt: 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
a, \(m_{H_2SO_4}=\frac{15\%\cdot156,8}{100\%}=23,52\left(g\right)\)
=>\(n_{H_2SO_4}=\frac{m}{M}=\frac{23,52}{98}=0,24\left(mol\right)\)
Theo pthh ta có: \(n_{Al}=\frac{2}{3}n_{H_2SO_4}=\frac{2}{3}\cdot0,24=0,16\left(mol\right)\)
=>\(m_{Al}=0,16.27=4,32\left(g\right)\)
b, \(m_{dd\left(sau.pứ\right)}=156,8+4,32-\left(0,24.2\right)=160,64\left(g\right)\)
\(C\%_{Al_2\left(SO4\right)_3}=\frac{\frac{1}{3}\cdot0,24.\left(27.2+32.3+16.4.3\right)}{160,64}\cdot100\%=\frac{27,36}{160,64}\cdot100\%=17,03\%\)
\(\text{Phản ứng: 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2}\)
a)
Ta có: m H2SO4=156,8.15%=23,52 gam
-> nH2SO4=23,52/98=0,24 mol
\(\text{-> Theo ptpu: nAl=2/3 nH2SO4=0,16 mol }\)
\(\text{-> mAl=0,16.27=4,32 gam}\)
b) Dung dịch sau phản ứng chứa Al2(SO4)3
Ta có: n Al2(SO4)3=1/2 n Al=0,08 mol
-> mAl2(SO4)3=27,36 gam\(\text{BTKL: m dung dịch sau phản ứng=m Al + m dung dịch axit -m H2}\)
\(\text{Ta có: n H2=3/2 n Al=0,24 mol }\)
\(\text{-> m dung dịch sau phản ứng=4,32+156,8-0,24.2=160,64 gam}\)
\(\text{-> % Al2(SO4)3=17,03% }\)
