\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.2..................................0.2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(\dfrac{2}{15}.............................0.2\)
\(m_{Al}=\dfrac{2}{15}\cdot27=3.6\left(g\right)\)
Thí nghiệm 1 :
\(Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)\\ \Rightarrow m_{tăng} = m_{Fe} -m_{H_2} = 11,2-0,2.2 = 10,8(gam)\)
Thí nghiệm 2 :
\(2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{Al} = a(mol) \to n_{H_2} = 1,5a(mol)\\ \Rightarrow m_{tăng} = m_{Al} -m_{H_2}\\ \Leftrightarrow 10,8 = 27a + 1,5a.2\\ \Leftrightarrow a = 0,45 \Rightarrow m = 0,45.27 = 12,15(gam)\)
