\(m_{H_2SO_4}=\frac{400.9,8}{100}=39,2\left(g\right)\rightarrow n_{H_2SO_4}=\frac{m}{M}=\frac{39,2}{98}=0,4\left(mol\right)\)
\(PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
(mol) 2 3 1 3
(mol) 4/15 0,4 2/15 0,4
\(m_{Al}=n.M=\frac{4}{15}.27=7,2\left(g\right)\)
\(V_{H_2}=n.22,4=0,4.22,4=8,96\left(l\right)\)
\(C_1:m_{Al_2\left(SO_4\right)_3}=n.M=\frac{2}{15}.342=45,6\left(g\right)\)
\(C_2:m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=m_{Al}+m_{H_2SO_4}-m_{H_2}=7,2+39,2-\left(0,4.2\right)=45,6\left(g\right)\)
a) 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
\(m_{H_2SO_4}=400\times9,8\%=39,2\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\frac{39,2}{98}=0,4\left(mol\right)\)
a) Theo pt: \(n_{Al}=\frac{2}{3}n_{H_2SO_4}=\frac{2}{3}\times0,4=\frac{4}{15}\left(mol\right)\)
\(\Rightarrow m_{Al}=\frac{4}{15}\times27=7,2\left(g\right)\)
b) Theo pt: \(n_{H_2}=n_{H_2SO_4}=0,4\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,4\times22,4=8,96\left(l\right)\)
c) Cách 1:
Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{3}n_{H_2SO_4}=\frac{1}{3}\times0,4=\frac{2}{15}\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=\frac{2}{15}\times342=45,6\left(g\right)\)
Cách 2:
\(m_{H_2}=2\times0,4=0,8\left(g\right)\)
Theo ĐL BTKL ta có:
\(m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)
\(\Leftrightarrow7,2+39,2=m_{Al_2\left(SO_4\right)_3}+0,8\)
\(\Leftrightarrow m_{Al_2\left(SO_4\right)_3}=45,6\left(g\right)\)
a) PTHH : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
b) \(m_{H_2SO_4}=\frac{400.98}{100}=39,2\left(g\right)\)
=> \(n_{H_2SO_4}=\frac{39,2}{98}=0,4mol\)
Theo PTHH : \(n_{Al}:n_{H_2SO_4}=2:3\Rightarrow n_{Al}=\frac{2}{3}n_{H_2SO_4}=\frac{2}{3}0,4=\frac{4}{15}mol\)
=> mAl = 7,2 g
c) Theo PTHH : \(n_{H_2}:n_{H_2SO_4}=1:1\Rightarrow n_{H_2}=n_{H_2SO_4}=0,4mol\)
=> VH2 = 0,4 . 22,4 = 8,96 l
d) C1 : Áp dụng định luật bảo toàn khối lượng ta có :
mAl + mH2SO4 = mAl2(SO4)3 + mH2
=> 7,2 + 39,2 = mAl2(SO4)3 + 0,8
=> mAl2(SO4)3 = 45,6 g
C2 : Theo PTHH : \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{3}n_{H_2SO_4}=\frac{1}{3}0,4=\frac{2}{15}mol\)
=> mAl2(SO4)3 = 2/15 . 342 = 45,6 g
