a)\ K_2CO_3+2HCl->2KCl+CO_2+H_2O b) n_{K_2CO_3}={13,8}/{138}=0,1(mol) Theo PT: n_{CO_2}=n_{K_2CO_3}=0,1(mol) ->V_{CO_2}=0,1.22,4=2,24(l) c) Theo PT: n_{KCl}=n_{HCl}=2n_{CO_2}=0,2(mol) ->C\%_{HCl}={0,2.36,5}/{200}.100\%=3,65\% d) m_{dd\ spu}=13,8+200-0,1.44=209,4(g) ->C\%_{KCl}={0,2.74,5}/{209,4}.100\%\approx 7,12\%
