Приготовить 25 мл хлорида калия молярная концентрация которого 0.2 моль/л​

1)2CH3-CHBr-CH2-CH3+2Na=CH3-CH2-CH-CH-CH2-CH3+2NaBr

                                                                              |        |

                                                                           CH3 CH3-3,4-диметилгексан.

                        CH3                            CH3 CH3

                          |                                      |     |

2)2CH3-CH2-C-J+2Na=CH3-CH2-C - C-CH2-CH3+2NaJ

                          |                                      |     |      

                       CH3                            CH3  CH3       3,3,4,4-тетраметилгексан.

                        200г 10%                        хг

BaCl2      +      H2SO4    =    2HCl    +    BasO4

                       v=1моль                          v=1моль

                       М=98г/моль                     М=233г/моль

                       m=98г                               m=233г

m(H2SO4)=(200г*10%)/100%=20г

20г/98г=хг/233г

х=(233г*20г)/98г=47,55г

ответ:47,55г

2)5%прим.             20%

  200г                    200г                                                 хл

    CaCO3      +     2HCl     =   CaCl2     +    H2O   +   CO2

  v=1моль             v=2моль                                        v=1моль

М=100г/моль        М=36,5г/моль                               Vm=22,4л/моль

m=100г                 m=73г                                             V=22,4л

w%(CaCO3)=100%-5%=95%(95% чистого вещества)

m(CacO3)=(95%*200г)/100%=190г

m(HCl)=(200г*20%)/100%=40г

190г/100г > 40г/73г

HCl-недостаток,по нему ведем рассчет

40г/73г=хл/22,4л

х=(22,4л*40г)/73г=12,27л

ответ:12,27л