1)

Phản ứng xảy ra:

\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)

Ta có:

\({n_{Zn}} = \frac{{6,5}}{{65}} = 0,1{\text{ mol}}\)

\({n_{{H_2}S{O_4}}} = \frac{{19,6}}{{98}} = 0,2{\text{ mol > }}{{\text{n}}_{Z{\text{n}}}}\)

Vậy \(H_2SO_4\) dư

\( \to {n_{{H_2}S{O_4}{\text{ dư}}}} = 0,2 - 0,1 = 0,1{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}{\text{ dư}}}} = 0,1.98 = 9,8{\text{ gam}}\)

\({n_{{H_2}}} = {n_{Zn}} = 0,1{\text{ mol}}\)

\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

2)
Phản ứng xảy ra:

\(F{{\text{e}}_2}{O_3} + 3{H_2}S{O_4}\xrightarrow{{}}F{{\text{e}}_2}{(S{O_4})_3} + 3{H_2}O\)

Ta có:

\({n_{F{{\text{e}}_2}{O_3}}} = \frac{{80}}{{56.2 + 16.3}} = 0,5{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = \frac{{98}}{{98}} = 1{\text{ mol}}\)

\({n_{{H_2}S{O_4}}} < 3{n_{F{{\text{e}}_2}{O_3}}}\)

Vậy \(Fe_2O_3\) dư

\( \to {n_{F{{\text{e}}_2}{O_3}{\text{ dư}}}} = 0,5 - \frac{1}{3} = \frac{1}{6}{\text{ mol}}\)

\( \to {m_{F{{\text{e}}_2}{O_3}}} = \frac{1}{6}.(56.2 + 16.3) = 26,67{\text{ gam}}\)

\({n_{F{{\text{e}}_2}{{(S{O_4})}_3}}} = \frac{1}{3}{n_{{H_2}S{O_4}}} = \frac{1}{3}{\text{ mol}}\)

\( \to {m_{F{{\text{e}}_2}{{(S{O_4})}_3}}} = \frac{1}{3}.(56.2 + 96.3) = 133,33{\text{ gam}}\)

3)

Phản ứng xảy ra:

\(Mg + {H_2}S{O_4}\xrightarrow{{}}MgS{O_4} + {H_2}\)

Ta có:

\({n_{{H_2}}} = \frac{{1,12}}{{22,4}} = 0,05{\text{ mol}}\)

Theo phản ứng:

\({n_{Mg}} = {n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,05{\text{ mol}}\)

\( \to {m_{Mg}} = 0,05.24 = 1,2{\text{ gam}}\)

\({C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{{n_{{H_2}S{O_4}}}}}{{{V_{dd}}}} = \frac{{0,05}}{{0,05}} = 1M\)

a) nFe=0,4(mol); nH2SO4=0,5(mol)

PTHH: Fe + H2SO4 -> FeSO4 + H2

Ta có: 0,4/1 < 0,5/1

=> Fe hết, H2SO4 dư. tính theo nFe.

=> nH2= nH2SO4(p.ứ)=nFe=0,4(mol)

=>nH2SO4(dư)=0,5-0,4=0,1(mol)

=>H2SO4(dư)=0,1.98=9,8(g)

b) V(H2,dktc)=0,4.22,4=8,96(l)