Ta có: \(TH_1:n_{Fe}=\dfrac{22,4}{22,4}=1\left(mol\right)\)
\(TH_2:n_{Fe}=\dfrac{224}{22,4}=10\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
Theo TH1: \(n_{HCl}=2.n_{Fe}=2.1=2\left(mol\right)\)
=> \(V_{dd_{HCl}}=\dfrac{2}{4}=0,5\left(lít\right)\)
Theo TH2: \(n_{HCl}=2.n_{Fe}=2.10=20\left(mol\right)\)
=> \(V_{dd_{HCl}}=\dfrac{20}{4}=5\left(lít\right)\)
Theo TH1: \(V_{dd_{FeCl_2}}=0,5\left(lít\right)\)
Theo PT: \(n_{FeCl_2}=n_{Fe}=1\left(mol\right)\)
=> \(C_{M_{FeCl_2}}=\dfrac{1}{0,5}=2M\)
Theo TH2: \(V_{dd_{FeCl_2}}=5\left(lít\right)\)
Theo PT: \(n_{FeCl_2}=n_{Fe}=10\left(g\right)\)
=> \(C_{M_{FeCl_2}}=\dfrac{10}{5}=2M\)
(Do đề cho số gam sắt ko rõ ràng nên mik ra 2 TH, TH nào đúng bn chép vào nhé.)
