a/
$nNO=\frac{4,48}{22,4}=0,2$
$Fe^0 \to Fe^{3+}+3e$
$Fe^{2+} \to Fe^{3+}+e$
$N^{+5}+3e \to N^{+2}$
$BTe⇒3nFe+nFe(OH)_2=3nNO=0,6(1)$
$mFe+mFe(OH)_2=32,6$
$⇒56nFe+90nFe(OH)_2=32,6(2)$
$(1)(2)\left \{ {{nFe=0,1} \atop {nFe(OH)_2=0,3}} \right.$
$\%mFe=\frac{0,1.56}{32,6}.100=17,18\%$
b/
$∑nFe(NO_3)_3=nFe+nFe(OH)_2=0,4$
$CMFe(NO_3)_3=\frac{0,4}{1}=0,4M$
$nHNO_3=1.1,6=1,6$
$OH^- +H^+ \to H_2O$
$∑nHNO_3 p/u=4nNO+2nFe(OH)_2=0,2.4+0,3.2=1,4$
$nHNO_3dư=1,6-1,4=0,2$
$CMHNO_3=\frac{0,2}{1}=0,2M$
c/
$OH^- +H^+ \to H_2O$
$Fe^{3+}+3OH^- \to Fe(OH)_3$
$∑nOH^-=nNaOH=0,2+0,4.3=1,4$
$VddNaOH=\frac{1,4}{0,75}=1,86lit$
d/
Gọi số mol $Fe(NO_3)_3$ p/u là $x$
$2Fe(NO_3)_3 \to Fe_2O_3+6NO_2+3/2O_2$
x 3x $\frac{3}{4}x$
$m_{giảm}=mNO_2+mO_2=32,4$
$⇔3x.46+\frac{3}{4}x.32=32,4$
$⇔x=0,2$
$H=\frac{0,2}{0,4}.100=50\%$