a)
Phản ứng xảy ra:
\(F{e_2}{(S{O_4})_3} + 6KOH\xrightarrow{{}}2Fe{(OH)_3} + 3{K_2}S{O_4}\)
Ta có:
\({n_{F{e_2}{{(S{O_4})}_3}}} = \frac{{40}}{{56.2 + 96.3}} = 0,1{\text{ mol;}}{{\text{n}}_{KOH}} = \frac{{56}}{{56}} = 1{\text{ mol > 6}}{{\text{n}}_{F{e_2}{{(S{O_4})}_3}}}\)
Do vậy KOH dư.
\( \to {n_{Fe{{(OH)}_3}}} = 2{n_{F{e_2}{{(S{O_4})}_3}}} = 0,2{\text{ mol;}}{{\text{n}}_{{K_2}S{O_4}}} = 3{n_{F{e_2}{{(S{O_4})}_3}}} = 0,3{\text{ mol; }}{{\text{n}}_{KOH{\text{ dư}}}} = 1 - 0,6 = 0,4{\text{ mol}}\)
BTKL:
\({m_{dd{\text{ sau phản ứng}}}} = {m_{F{e_2}{{(S{O_4})}_3}}} + {m_{KOH}} - {m_{Fe{{(OH)}_3}}} = 40 + 56 - 0,2.(56 + 17.3) = 74,6{\text{ gam}}\)
Ta có:
\({m_{{K_2}{{(S{O_4})}_3}}} = 0,3.(39.2 + 96) = 52,2{\text{ gam;}}{{\text{m}}_{KOH{\text{ dư}}}} = 0,4.56 = 22,4{\text{ gam}}\)
\( \to \% {m_{{K_2}S{O_4}}} = \frac{{52,2}}{{74,6}} = 70\% \to \% {m_{KOH{\text{ dư}}}} = 30\% \)
b)
Phản ứng xảy ra:
\(MgS{O_4} + 2NaOH\xrightarrow{{}}Mg{(OH)_2} + N{a_2}S{O_4}\)
Ta có:
\({m_{MgS{O_4}}} = 120.10\% = 12{\text{ gam}} \to {n_{MgS{O_4}}} = \frac{{12}}{{120}} = 0,1{\text{ mol;}}{{\text{m}}_{NaOH}} = 80.5\% = 4{\text{ gam}} \to {{\text{n}}_{NaOH}} = 0,1{\text{ mol < 2}}{{\text{n}}_{MgS{O_4}}}\)
\( \to {n_{Mg{{(OH)}_2}}} = \frac{1}{2}{n_{NaOH}} = 0,05{\text{ mol;}}{{\text{n}}_{N{a_2}S{O_4}}} = \frac{1}{2}{n_{NaOH}} = 0,05{\text{ mol;}}{{\text{n}}_{MgS{O_4}{\text{ dư}}}} = 0,1 - 0,05 = 0,05{\text{ mol}}\)
BTKL: \({m_{dd{\text{sau phản ứng}}}} = {m_{ddMgS{O_4}}} + {m_{dd{\text{ NaOH}}}} - {m_{Mg{{(OH)}_2}}} = 120 + 80 - 0,05.(24 + 17.2) = 197,1{\text{ gam}}\)
\({m_{N{a_2}S{O_4}}} = 0,05.(23.2 + 96) = 7,1{\text{ gam;}}{{\text{m}}_{MgS{O_4}}} = 6{\text{ gam}}\)
\( \to C{\% _{N{a_2}S{O_4}}} = \frac{{7,1}}{{197,1}} = 3,6\% ;{\text{ C}}{{\text{\% }}_{MgS{O_4}}} = \frac{6}{{197,1}} = 3,04\% \)
