HCl+NH3=NH4Cl согласно уравнения количество аммиака вступившего в реакцию равно количеству хлористого водорода, соответственно их объёмы будут равны VHCl=4,48литра.
"...позитивную йодоформную пробу..." - позитивно звучит)
Исходное вещество С5Н12О - 3-метилбутанол-2.
Продукт окисления C5H10O - 3-метилбутанон.
Алкен C5H10 - 2-метилбутен-2.
Запишемо рівняння реакції
![NH_3+HCl\to NH_4Cl](https://tex.z-dn.net/?f=NH_3%2BHCl%5Cto+NH_4Cl)
Знайдемо масу амоній хлорида
![\,\,\,\,6.8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\\ NH_3+HCl\to NH_4Cl \\ 17\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,53.5](https://tex.z-dn.net/?f=%5C%2C%5C%2C%5C%2C%5C%2C6.8%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2Cx%5C%5C+NH_3%2BHCl%5Cto+NH_4Cl+%5C%5C+17%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C+%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C53.5)
![m(NH_4Cl)= \dfrac{6.8\cdot53.5}{17} =21.4](https://tex.z-dn.net/?f=m%28NH_4Cl%29%3D+%5Cdfrac%7B6.8%5Ccdot53.5%7D%7B17%7D+%3D21.4)
грам
Знаходимо масу розчину: m(р-ну)=m(NH₄Cl)+m(H₂O)=21.4+200=221.4 г
Масова частка NH4Cl тоді
![w(NH_4Cl)= \dfrac{m(p-pHu)}{m(p-pHy)} \cdot100= \dfrac{21.4}{221.4} \cdot100\approx10](https://tex.z-dn.net/?f=w%28NH_4Cl%29%3D+%5Cdfrac%7Bm%28p-pHu%29%7D%7Bm%28p-pHy%29%7D+%5Ccdot100%3D+%5Cdfrac%7B21.4%7D%7B221.4%7D+%5Ccdot100%5Capprox10)
Окончательный ответ: w(NH₄Cl)=10%