a)
Ta có:
\({M_{N{a_2}C{O_3}.10{H_2}O}} = 23.2 + 12 + 16.3 + 18.10 = 286\)
\( \to \% {m_{{H_2}O}} = \frac{{18.10}}{{286}} = 62,94\% \)
b)
Ta có:
\({M_{CaS{O_4}.2{H_2}O}} = 40 + 32 + 16.4 + 18.2 = 172\)
\( \to \% {m_{{H_2}O}} = \frac{{18.2}}{{172}} = 20,93\% \)