Bài 8:
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1--->0,15------>0,05---------->0,15
a. \(V_{H_2SO_4}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\)
b. \(V'=0,15.22,4=3,36\left(l\right)\)
c. \(CM_{Al_2\left(SO_4\right)_3}=\dfrac{0,05}{0,15}=\dfrac{1}{3}M\)
Bài 9:
\(n_{Fe}=\dfrac{2,8}{56}=0,2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2-->0,4---->0,2------>0,2
a. \(V_{HCl}=\dfrac{0,4}{1}=0,4\left(l\right)=400\left(ml\right)\)
b.\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c. \(CM_{FeCl_2}=\dfrac{0,2}{0,4}=0,5M\)
Bài 10:
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1<--0,2<----0,1<----0,1
a. \(m=m_{Fe}=0,1.56=5,6\left(g\right)\)
b. \(V_{HCl}=\dfrac{0,2}{0,1}=2\left(l\right)=200\left(ml\right)\)
c.\(CM_{FeCl_2}=\dfrac{0,1}{2}=0,05M\)