a)
Để NaOH phản ứng tối thiểu thì phản ứng xảy ra như sau:
\(NaOH + S{O_2}\xrightarrow{{}}NaHS{O_3}\)
Ta có:
\({n_{S{O_2}}} = {n_{NaOH}} = \frac{{0,112}}{{22,4}} = 0,005{\text{ mol}} \to {{\text{V}}_{NaOH}} = \frac{{0,005}}{2} = 0,0025{\text{ lít = 2}}{\text{,5 ml}}\)
b)
Ta có:
\({n_{Ba{{(OH)}_2}}} = 0,1.1,5 = 0,15{\text{ mol > }}{{\text{n}}_{BaC{O_3}}} = \frac{{19,7}}{{197}} = 0,1{\text{ mol}}\)
Do vậy có 2 trường hợp xảy ra:
TH1: \(Ba{(OH)_2}\) dư.
\(Ba{(OH)_2} + C{O_2}\xrightarrow{{}}BaC{O_3} + {H_2}O\)
\( \to {n_{C{O_2}}} = {n_{BaC{O_3}}} = 0,1{\text{ mol}} \to {\text{V = 0}}{\text{,1}}{\text{.22}}{\text{,4 = 2}}{\text{,24 lít}}\)
TH2: \(Ba{(OH)_2}\) hết
\(Ba{(OH)_2} + C{O_2}\xrightarrow{{}}BaC{O_3} + {H_2}O\)
\(Ba{(OH)_2} + 2C{O_2}\xrightarrow{{}}Ba{(HC{O_3})_2}\)
\({n_{BaC{O_3}}} = 0,1{\text{ mol}} \to {{\text{n}}_{Ba{{(HC{O_3})}_2}}} = 0,15 - 0,1 = 0,05{\text{ mol}}\)
\( \to {n_{C{O_2}}} = {n_{BaC{O_3}}} + 2{n_{Ba{{(HC{O_3})}_2}}} = 0,1 + 0,05.2 = 0,2{\text{ mol}}\)
\( \to V = 0,2.22,4 = 4,48{\text{ lít}}\)
