\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,1 0,3
\(m_{Al}=0,2.27=5,4\left(g\right)\)
\(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4.100}{19,6}=150\left(g\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,1.100}{5,4+150-0,3.2}=22,09\%\)
a) \(2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\)
PT: 2 3 1 3 (mol)
ĐB: 0,2 0,3 0,1 0,3 (mol)
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
b) \(m_{Al}=n.M=0,2.27=5,4\left(g\right)\)
c) \(m_{H_2SO_4}=n.M=0,3.98=29,4\left(g\right)\)
\(m_{dd_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{C\%}.100\%=\dfrac{29,4}{19,6\%}.100\%=150\left(g\right)\)
d) \(m_{Al_2\left(SO_4\right)_3}=n.M=0,1.342=34,2\left(g\right)\)
\(m_{H_2}=n.M=0,3.2=0,6\left(g\right)\)
\(m_{dd_{Al_2\left(SO_4\right)_3}}=m_{Al}+m_{dd_{H_2SO_4}}-m_{H_2}=5,4+150-0,6=154,8g\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{m_{Al_2\left(SÔ_4\right)_3}}{m_{dd_{Al_2\left(SÔ_4\right)_3}}}.100\%=\dfrac{34,2}{154,8}.100\%=22,1\%\)