Đáp án:
${m_ \downarrow } = 5\,\,gam$
Giải thích các bước giải:
${n_{C{O_2}}} = 0,1\,\,mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$
${n_{Ca{{(OH)}_2}}} = 0,05\,\,mol;\,\,{n_{KOH}} = 0,2\,\,mol$
$\to {n_{O{H^ - }}} = 0,05.2 + 0,2 = 0,3\,\,mol$
Xét: $\dfrac{{{n_{O{H^ - }}}}}{{{n_{C{O_2}}}}} = 3 \to tạo\,\,muối\,\,CO_3^{2 - }$
${n_{CO_3^{2 - }}} = {n_{C{O_2}}} = 0,1\,\,mol$
PTHH:
$C{a^{2 + }} + CO_3^{2 - } \to CaC{O_3} \downarrow $
$0,05\,\,\,\,\,\,\,\,0,1 \to \,\,\,\,\,\,\,0,05\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol$
$\to {m_ \downarrow } = 0,05.100 = 5\,\,gam$
