Câu 11:

nFe3O4=$\frac{23,2}{232}$ =0,1 mol

Fe3O4+4H2SO4→FeSO4+Fe2(SO4)3+4H2O

 0,1→                      0,1           0,1

m dd sau=m Fe3O4 +m H2SO4

               =23,2+840=863,2 g

m ct FeSO4=0,1 x 152 = 15,2 g

m ct Fe2(SO4)3= 0,1 x 400=40 g

C% FeSO4=$\frac{15,2}{863,2}$ x100≈1,76 %

C% Fe2(SO4)3=$\frac{40}{863,2}$ x100≈4,6 %

Câu 12 : chịu

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11)

Phản ứng xảy ra:

\(F{e_3}{O_4} + 4{H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + F{e_2}{(S{O_4})_3} + 4{H_2}O\)

Ta có: 

\({n_{F{e_3}{O_4}}} = \frac{{23,2}}{{232}} = 0,1{\text{ mol;}}{{\text{V}}_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{840}}{{1,4}} = 600ml = 0,6{\text{ lít}} \to {{\text{n}}_{{H_2}S{O_4}}} = 0,6.2 = 1,2{\text{ mol}}\)

Vì \({n_{{H_2}S{O_4}}} > 4{n_{F{e_3}{O_4}}}\) nên axit dư.

\( \to {n_{FeS{O_4}}} = {n_{F{e_2}{{(S{O_4})}_3}}} = {n_{F{e_3}{O_4}}} = 0,1{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = 1,2 - 4{n_{F{e_3}{O_4}}} = 0,8{\text{ mol}}\)

\( \to {m_{FeS{O_4}}} = 15,2{\text{ gam;}}{{\text{m}}_{F{e_2}{{(S{O_4})}_3}}} = 40{\text{ gam;}}{{\text{m}}_{{H_2}S{O_4}{\text{ dư}}}} = 78,4{\text{ gam}}\)

\({m_{dd}} = 23,2 + 840 = 863,2{\text{ gam}}\)

\( \to C{\% _{FeS{O_4}}} = 1,76\% ;C{\% _{F{e_2}{{(S{O_4})}_3}}} = 4,63\% ;C{\% _{{H_2}S{O_4}}} = 9,08\% \)

12)

Phản ứng xảy ra:

\(CaO + {H_2}O\xrightarrow{{}}Ca{(OH)_2}\)

Gọi \({n_{CaO}} = x \to {m_{CaO}} = 56x{\text{ gam}} \to {{\text{m}}_{{H_2}O}} = 56x.140\%  = 78,4x{\text{gam}}\)

Ta có: 

\({n_{CaO}} = {n_{{H_2}O{\text{ phản ứng}}}} = x \to {m_{{H_2}O{\text{ phản ứng}}}} = 18x \to {m_{{H_2}O{\text{ dư}}}} = 78,4 - 18x = 60,4x{\text{ gam}}\)

\( \to \% {H_2}O{\text{ dư = }}\frac{{60,4x}}{{78,4x}} = 77,04\% \)