Ta có: \(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)
PTHH: Fe2O3 + 6HNO3 ---> 2Fe(NO3)3 + 3H2O
Theo PT: \(n_{HNO_3}=6.n_{Fe_2O_3}=6.0,3=1,8\left(mol\right)\)
=> \(m_{HNO_3}=1,8.63=133,4\left(g\right)\)
Ta có: \(\dfrac{133,4}{m_{dd_{HNO_3}}}.100\%=12,6\%\)
=> \(m_{dd_{HNO_3}}\approx1059\left(g\right)\)
=> \(m_{dd_{Fe\left(NO_3\right)_3}}=48+1059=1107\left(g\right)\)
Theo PT: \(n_{Fe\left(NO_3\right)_3}=2.n_{Fe_2O_3}=2.0,3=0,6\left(mol\right)\)
=> \(m_{Fe\left(NO_3\right)_3}=0,6.242=145,2\left(g\right)\)
=> \(C\%_{Fe\left(NO_3\right)_3}=\dfrac{145,2}{1107}.100\%\approx13,12\%\)
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