1)
Phản ứng xảy ra:
\({H_2}S{O_4} + 2NaOH\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
\({H_2}S{O_4} + 2KOH\xrightarrow{{}}{K_2}S{O_4} + 2{H_2}O\)
Ta có:
\({n_{{H_2}S{O_4}}} = 0,02.1 = 0,02{\text{ mol}} \to {{\text{n}}_{NaOH}} = {n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,04{\text{ mol}}\)
Ta có:
\({m_{NaOH}} = 0,04.40 = 1,6{\text{ gam}} \to {{\text{m}}_{dd{\text{ NaOH}}}} = \frac{{1,6}}{{20\% }} = 8{\text{ gam}} \to {{\text{V}}_{dd{\text{NaOH}}}} = \frac{8}{{1,2}} = 6,67ml\)
\({m_{KOH}} = 0,04.56 = 2,24{\text{ gam}} \to {{\text{m}}_{dd{\text{ KOH}}}} = \frac{{2,24}}{{9,6\% }} = 23,33gam \to {V_{dd{\text{ KOH}}}} = \frac{{23,33}}{{1,045}} = 22,32ml\)
2)
Phản ứng xảy ra:
\(2R{(OH)_3} + 3{H_2}S{O_4}\xrightarrow{{}}{R_2}{(S{O_4})_3} + 3{H_2}O\)
Ta có:
\({n_{{H_2}S{O_4}}} = 0,1.0,75 = 0,075{\text{ mol}} \to {{\text{n}}_{R{{(OH)}_3}}} = \frac{2}{3}{n_{{H_2}S{O_4}}} = 0,05{\text{ mol}} \to {{\text{M}}_{R{{(OH)}_3}}} = R + 17.3 = \frac{{0,9}}{{0,05}} = 18 \to R < 0\)
Câu này sai nhé.