3)

a)

Ta có:

\({m_{chất{\text{tan}}}} = {m_{Ca{{(OH)}_2}}} = 7,4{\text{ gam}}\)

\({m_{dd}} = {m_{Ca{{(OH)}_2}}} + {m_{{H_2}O}} = 7,4 + 20 = 27,4{\text{ gam}}\)

b)

\({m_{chất\;{\text{tan}}}} = {m_{HCl}} = 7,3{\text{ gam}}\)

\({m_{dd}} = {m_{HCl}} + {m_{{H_2}O}} = 7,3 + 20 = 27,3{\text{ }}gam\)

c)

Phản ứng xảy ra:

\({K_2}O + {H_2}O\xrightarrow{{}}2KOH\)

\({n_{{K_2}O}} = \frac{{9,4}}{{39.2 + 16}} = 0,1{\text{ mol}} \to {{\text{n}}_{KOH}} = 2{n_{{K_2}O}} = 0,2{\text{ mol}}\)

\( \to {m_{chất{\text{ tan}}}} = {m_{KOH}} = 0,2.56 = 11,2{\text{ gam}}\)

\({m_{dd}} = {m_{{K_2}O}} + {m_{{H_2}O}} = 9,4 + 90 = 99,4{\text{ gam}}\)

e)

Phản ứng xảy ra:

\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)

Ta có:

\({n_{Na}} =n_{NaOH}= \frac{{2,3}}{{23}} = 0,1{\text{ mol}}\)

\( \to {n_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,05{\text{ mol}}\)

\({m_{chất{\text{ tan}}}} = {m_{NaOH}} = 0,1.40 = 4{\text{ gam}}\)

\({m_{dd}} = {m_{Na}} + {m_{{H_2}O}} - {m_{{H_2}}} = 2,3 + 80 - 0,05.2 = 82,2{\text{ gam}}\)

f)

Phản ứng xảy ra:

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

\({n_{Zn}} = \frac{{6,5}}{{65}} = 0,1{\text{ mol = }}{{\text{n}}_{ZnC{l_2}}} = {n_{{H_2}}}\)

\( \to {m_{chất{\text{ tan}}}} = {m_{ZnC{l_2}}} = 0,1.(65 + 35,5.2) = 13,6{\text{ gam}}\)

BTKL:

\({m_{Zn}} + {m_{dd\;{\text{HCl}}}} = {m_{dd\;}} + {m_{{H_2}}}\)

\( \to {m_{dd}} = 6,5 + 70 - 0,1.2 = 76,3{\text{ gam}}\)

4)

a)

Ta có:

\({M_{N{a_2}C{O_3}.10{H_2}O}} = 23.2 + 12 + 16.3 + 18.10 = 286\)

\( \to \% {m_{{H_2}O}} = \frac{{18.10}}{{286}}.100\%  = 62,937\% \)

b)

\({M_{CaS{O_4}.2{H_2}O}} = 40 + 32 + 16.4 + 18.2 = 172\)

\( \to \% {m_{{H_2}O}} = \frac{{18.2}}{{172}}.100\%  = 20,93\% \)

5)

Ta có:

\({n_{CuS{O_4}.5{H_2}O}} = \frac{1}{{64 + 32 + 18.5}} = \frac{1}{{250}}{\text{ kmol = }}{{\text{n}}_{CuS{O_4}}}\)

\( \to {m_{CuS{O_4}}} = \frac{1}{{250}}.(64 + 32 + 16.4) = 0,64{\text{ kg}}\)

6)

Gọi công thức của muối ngậm nước có dạng \(RSO_4.nH_2O\)

Ta có:

\({M_{RS{O_4}.n{H_2}O}} = {M_R} + 32 + 16.4 + 18n = {M_R} + 96 + 18n\)

\( \to \% {m_S} = \frac{{32}}{{{M_R} + 96 + 18n}} = 11,51\% \)

\( \to {M_R} + 96 + 18n = 278\)

\(\% {m_{{H_2}O}} = \frac{{18n}}{{{M_R} + 96 + 18n}} = \frac{{18n}}{{278}} = 45,324\%  \to n = 7\)

\( \to {M_R} = 56 \to R:F{\text{e}}\)