Question

cho m(g) fe tác dụng vừa đủ với 200g dd HCl 14,6% . tìm m

Answer 1

\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{1}{2}n_{HCl}=0,4\left(mol\right)\\ \Rightarrow m_{Fe}=0,4.56=22,4\left(g\right)\)

Answer 2

nHCl=200.14,6%36,5=0,8(mol)Fe+2HCl→FeCl2+H2nFe=12nHCl=0,4(mol)⇒mFe=0,4.56=22,4(g)