Đáp án:
\( \to m = 23,3{\text{ gam}}\)
\({m_{Ba{{(OH)}_2}}} = 17,1{\text{ gam}}\)
\({m_{BaC{O_3}}} = 19,7{\text{ gam}}\)
\({m_{C{O_2}}} = 2,24{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(BaO + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} + {H_2}O\)
\(BaO + {H_2}O\xrightarrow{{}}Ba{(OH)_2}\)
Ta có:
\({n_{BaO}} = \frac{{30,6}}{{137 + 16}} = 0,2{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = \frac{{9,8}}{{98}} = 0,1{\text{ mol}}\)
\( \to {n_{BaS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,1{\text{ mol;}}{{\text{n}}_{Ba{{(OH)}_2}}} = 0,2 - 0,1 = 0,1{\text{ mol}}\)
\( \to m = {m_{BaS{O_4}}} = 0,1.(137 + 96) = 23,3{\text{ gam}}\)
\({m_{Ba{{(OH)}_2}}} = 0,1.(137 + 17.2) = 17,1{\text{ gam}}\)
Cho dung dịch \(A\) tác dụng với \(CO_2\)
\(Ba{(OH)_2} + C{O_2}\xrightarrow{{}}BaC{O_3} + {H_2}O\)
\( \to {n_{BaC{O_3}}} = {n_{C{O_2}}} = {n_{Ba{{(OH)}_2}}} = 0,1{\text{ mol}}\)
\( \to {m_{BaC{O_3}}} = 0,1.(137 + 60) = 19,7{\text{ gam}}\)
\({m_{C{O_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
