Fe + CuCl2 → FeCl2 + Cu (1)
\(n_{Fe}=\frac{2,8}{56}=0,05\left(mol\right)\)
\(n_{CuCl_2}=0,04\times1=0,04\left(mol\right)\)
Theo pT1: \(n_{Fe}=n_{CuCl_2}\)
Theo bài: \(n_{Fe}=\frac{5}{4}n_{CuCl_2}\)
Vì \(\frac{5}{4}>1\) ⇒ Fe dư
a) Chất rắn A: Fe dư và Cu
Theo pT1: \(n_{Cu}=n_{CuCl_2}=0,04\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,04\times64=2,56\left(g\right)\)
Theo pT1: \(n_{Fe}pư=n_{CuCl_2}=0,04\left(mol\right)\)
\(\Rightarrow n_{Fe}dư=0,05-0,04=0,01\left(mol\right)\)
\(\Rightarrow m_{Fe}dư=0,01\times56=0,56\left(g\right)\)
\(\Rightarrow m_A=2,56+0,56=3,12\left(g\right)\)
b) FeCl2 + 2NaOH → 2NaCl + Fe(OH)2↓ (2)
Theo PT1: \(n_{FeCl_2}=n_{CuCl_2}=0,04\left(mol\right)\)
Theo Pt2: \(n_{NaOH}=2n_{FeCl_2}=2\times0,04=0,08\left(mol\right)\)
\(\Rightarrow V_{ddNaOH}=\frac{0,08}{5}=0,016\left(l\right)=16\left(ml\right)\)
c) Cu + 4HNO3 → Cu(NO3)2 + 2NO2↑ + 2H2O (3)
Fe + 4HNO3 → Fe(NO3)3 + NO2↑ + 2H2O (4)
Theo PT3: \(n_{NO_2}=2n_{Cu}=2\times0,04=0,08\left(mol\right)\)
Theo PT4: \(n_{NO_2}=n_{Fe}dư=0,01\left(mol\right)\)
\(\Rightarrow\Sigma n_{NO_2}=0,08+0,01=0,09\left(mol\right)\)
\(\Rightarrow V_{NO_2}=0,09\times22,4=2,016\left(l\right)\)