1)
Phản ứng xảy ra:
\(Fe + 3AgN{O_3}\xrightarrow{{}}Fe{(N{O_3})_3} + 3Ag\)
Ta có:
\({n_{Fe}} = \frac{{16,8}}{{56}} = 0,3{\text{ mol}}\)
\( \to {n_{Ag}} = 3{n_{Fe}} = 0,3.3 = 0,9{\text{ mol}}\)
\( \to {m_{Ag}} = 0,9.108 = 97,2{\text{ gam}}\)
2)
Ta có:
\({n_{Fe}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol;}}{{\text{n}}_{AgN{O_3}}} = 0,5.1 = 0,5{\text{ mol}}\)
\( \to \frac{{{n_{AgN{O_3}}}}}{{{n_{Fe}}}} = \frac{{0,5}}{{0,2}} = 2,5\)
Do vậy xảy ra 2 phản ứng
\(Fe + 3AgN{O_3}\xrightarrow{{}}Fe{(N{O_3})_3} + 3Ag\)
\(Fe + 2AgN{O_3}\xrightarrow{{}}Fe{(N{O_3})_2} + 2Ag\)
\( \to {n_{Ag}} = {n_{AgN{O_3}}} = 0,5{\text{ mol}}\)
\( \to m = {m_{Ag}} = 0,5.108 = 54{\text{ gam}}\)
Vì 2,5 là trung bình của 2 và 3.
\( \to {n_{Fe{{(N{O_3})}_2}}} = {n_{Fe{{(N{O_3})}_3}}} = \frac{1}{2}{n_{Fe}} = 0,1{\text{ mol}}\)
\({V_{dd}} = 500{\text{ ml = 0}}{\text{,5 lít}}\)
\( \to {C_{M{\text{ Fe(N}}{{\text{O}}_3}{)_2}}} = {C_{M{\text{ Fe(N}}{{\text{O}}_3}{)_3}}} = \frac{{0,1}}{{0,5}} = 0,2M\)
