1. Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
a, \(n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\Rightarrow m_{FeSO_4}=0,1.152=15,2\left(g\right)\)
b, \(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\Rightarrow C_{M_{FeSO_4}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)
2. \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
b, \(n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=0,15\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
\(\Rightarrow a=C\%_{H_2SO_4}=\dfrac{14,7}{98}.100\%=15\%\)
