10)
Ta có:
\({n_{{H_2}S{O_4}}} = 0,035.2 = 0,07{\text{ mol}}\)
Phản ứng xảy ra:
\(Ba{(OH)_2} + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} + 2{H_2}O\)
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
Ta có:
\({n_{BaS{O_4}}} = \frac{{9,32}}{{233}} = 0,04{\text{ mol = }}{{\text{n}}_{Ba{{(OH)}_2}}} \to {n_{NaOH}} = 2.(0,07 - 0,04).2 = 0,06{\text{ mol}}\)
\( \to {V_A} = {V_B} = \frac{{100}}{2} = 50{\text{ ml = 0}}{\text{,05 lít}}\)
\( \to {C_{M{\text{Ba(OH}}{{\text{)}}_2}}} = \frac{{0,04}}{{0,05}} = 0,8M;{C_{M{\text{ NaOH}}}} = \frac{{0,06}}{{0,05}} = 1,2M\)
Hòa tan 1,08 gam Al
Ta có:
\({n_{Al}} = \frac{{1,08}}{{27}} = 0,04{\text{ mol}}\)
\(Ba{(OH)_2} + 2Al + 2{H_2}O\xrightarrow{{}}Ba{(Al{O_2})_2} + 3{H_2}\)
\(2NaOH + 2Al + 2{H_2}O\xrightarrow{{}}2NaAl{O_2} + 3{H_2}\)
Ta có:
\({n_{NaOH}} = 0,02.0,6 = 0,012{\text{ mol}} \to {{\text{n}}_{Al}} = {n_{NaOH}} + 2{n_{Ba{{(OH)}_2}}} = 0,04 \to {n_{Ba{{(OH)}_2}}} = 0,014{\text{ mol}} \to {{\text{V}}_B} = \frac{{0,014}}{{0,8}} = 0,0175{\text{ lít = 17}}{\text{,5ml}}\)
11)
\({H_2}S{O_4} + 2KOH\xrightarrow{{}}{K_2}S{O_4} + 2{H_2}O\)
Vì để trung hòa 100 ml E cần 0,04 mol axit sunfuric
Suy ra 100 ml E chứa 0,08 mol KOH dư, vậy 500ml E chứa 0,4 mol KOH dư.
\( \to 0,3y - 0,2x.2 = 0,4\)
Trộn 300 ml C với 200 ml D
\({n_{{H_2}S{O_4}}} = 0,3x;{n_{KOH}} = 0,2y\)
TH1: Nếu F chứa axit dư
\(A{l_2}{O_3} + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}O\)
\({n_{A{l_2}{O_3}}} = \frac{{2,04}}{{102}} = 0,02{\text{ mol}} \to {{\text{n}}_{{H_2}S{O_4}}} = 0,02.3 = 0,06{\text{ mol}} \to {{\text{n}}_{{H_2}S{O_4}{\text{ trong 500ml F}}}} = 0,06.5 = 0,3{\text{ mol}}\)
\( \to 0,3x - \frac{{0,2y}}{2} = 0,3\)
Giải được: x=2,6; y=4,8.
TH2: Nếu F chứa KOH dư
\(2KOH + 2Al + 2{H_2}O\xrightarrow{{}}2KAl{O_2} + 3{H_2}O\)
\({n_{A{l_2}{O_3}}} = 0,02{\text{ mol}} \to {{\text{n}}_{KOH}} = 0,02.2 = 0,04{\text{ mol}} \to {{\text{n}}_{KOH{\text{ trong 500ml F}}}} = 0,04.5 = 0,2{\text{ mol}}\)
\( \to 0,2y - 0,3x.2 = 0,2\)
Giải được: x=16; y=5.
