Fe + 4HNO3 →Fe(NO3)3 + NO + 2H2O
\(n_{HNO_3}=0,4.2=0,8\left(mol\right)\)
TH1 : 2 chất đều phản ứng hết
\(n_{Fe}=\dfrac{0,8}{4}=0,2\left(mol\right)\)
=> m Fe=11,2g
\(n_{Fe\left(NO_3\right)_3}=\dfrac{0,8}{4}=0,2\left(mol\right)\)
=> \(m_{Fe\left(NO_3\right)_3}=0,2.242=48,4\left(g\right)\)
Theo đề m gam Fe ------> 3m gam muối
Mà \(\dfrac{48,4}{11,2}=4,32\)
Vậy TH này loại
TH2: HNO3 dư
Fe + 4HNO3 →Fe(NO3)3 + NO + 2H2O
\(\dfrac{m}{56}\)--->\(\dfrac{m}{14}\)---->\(\dfrac{m}{56}\)
=> \(\dfrac{m}{56}=\dfrac{3m}{56+62.3}\)
=> m=0 (loại)
TH3: Fe dư
Fe + 4HNO3 → Fe(NO3)3 + NO↑ + 2H2O
0,2<--0,8---------->0,2
Fe + 2Fe(NO3)3 → 3Fe(NO3)2
\(\dfrac{m}{56}-0,2\)---------------------------------------->\(3\left(\dfrac{m}{56}-0,2\right)\)
=> \(n_{Fe\left(NO_3\right)_3}=3\left(\dfrac{m}{56}-0,2\right)=\dfrac{3m}{56+62.3}\)
=>m=\(\dfrac{6776}{465}\left(g\right)\approx14,572\left(g\right)\)