Đáp án:

\(N{a_2}C{O_3}.10{H_2}O\)

Giải thích các bước giải:

 Bảo toàn nguyên tố X:

\({n_{CaX{O_3}}} = {n_{BaX{O_3}}} \to \frac{{2,5}}{{40 + X + 16.3}} = \frac{{4,925}}{{137 + X + 16.3}} \to X = 12 \to X:C\)

\( \to {n_{CaC{O_3}}} = \frac{{2,5}}{{100}} = 0,025{\text{ mol = }}{{\text{n}}_{C{O_3}^{2 - }}}\)

Trong tinh thể

\({m_{{H_2}O}} = 7,15.62,94\%  = 4,5{\text{ gam}} \to {{\text{n}}_{{H_2}O}} = \frac{{4,5}}{{18}} = 0,25{\text{ mol; }}{{\text{m}}_{{M_a}{{(C{O_3})}_b}}} = 7,15 - 4,5 = 2,65{\text{ gam}}\)

\({n_{{M_a}{{(C{O_3})}_b}}} = \frac{{0,025}}{b} \to M.a + 60b = \frac{{2,65}}{{\frac{{0,025}}{b}}} = 106b \to M = \frac{{46b}}{a} \to a = 2;b = 1 \to M = 23 \to Na\)

\( \to {n_{N{a_2}C{O_3}}} = 0,025{\text{ mol}} \to {\text{n = }}\frac{{0,25}}{{0,025}} = 10 \to N{a_2}C{O_3}.10{H_2}O\)