Question

!!!!Пожалуйста, помогите, нужно срочно!!!! Нагревали 44,5 жира (глицерида одной кислоты) с 70 мл 20% раствором гидроксида натрия (p = 1,2 г/см^3). Для нейтрализации избытка гидроксида натрия потребовалось 22,5 мл 36,5%-ой соляной кислоты (p = 1,2 г/см^3). Какие органические соединения и в каким количествах (по массе) получилось при этом из жира?

Answer 1

Ответ:

Глицерина 4,6 грамма

Стеарат натрия 45,9 грамма

Объяснение:

n(HCl)=(22.5*1.2*0.365)/36.5=0,27 моль

n(NaOH)=(70*1.2*0.2)/40=0,42 моль

nOst(NaOH)=(0.42-0.27)=0,15 моль

M(Tg)=44.5/((0.15/3))=890 г/моль

Такой молярной массе соответствует тристеароилглицерин.

В результате реакции получается 3 моль стеарата натрия и 1 моль глицерина в соотношение.

n(C57H110O6)=44.5/890=0,05 моль

m(C3H8O3)=0.05*92=4,6 грамма

m(C18H35O2Na)=0,05*3*306=45,9 грамм

Так, закон сохранения масс

44,5+6=4,6+45,9

50,5=50,5

Ну да, верно.

Answer 2

Calculation of the Amount of Organic Compounds Produced

To determine the amount of organic compounds produced from the given reaction, we need to calculate the moles of each reactant and product involved. From the balanced chemical equation, we can see that the reaction involves the neutralization of excess sodium hydroxide (NaOH) with hydrochloric acid (HCl) formed from the reaction between the fat and sodium hydroxide.

The balanced chemical equation for the reaction is as follows:

Fat + NaOH → Soap + Glycerol

Given: - Mass of fat (glyceride of one acid) = 44.5 g - Volume of 20% NaOH solution = 70 mL - Volume of 36.5% HCl solution = 22.5 mL

To calculate the moles of each reactant and product, we need to convert the given mass and volume into moles using the molar mass and molar concentration, respectively.

1. Moles of NaOH: - Molar concentration of NaOH = 20% = 20 g/100 mL - Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol - Volume of NaOH solution = 70 mL = 0.07 L - Moles of NaOH = (Molar concentration of NaOH * Volume of NaOH solution) / Molar mass of NaOH

2. Moles of HCl: - Molar concentration of HCl = 36.5% = 36.5 g/100 mL - Molar mass of HCl = 1.01 g/mol + 35.45 g/mol = 36.46 g/mol - Volume of HCl solution = 22.5 mL = 0.0225 L - Moles of HCl = (Molar concentration of HCl * Volume of HCl solution) / Molar mass of HCl

3. Moles of Fat (Glyceride of One Acid): - Mass of fat = 44.5 g - Molar mass of fat = ?? g/mol (not provided in the question)

Unfortunately, the molar mass of the fat (glyceride of one acid) is not provided in the question. Without this information, we cannot accurately calculate the moles of fat or determine the amount of organic compounds produced. To proceed further, we would need the molar mass of the fat.

Please provide the molar mass of the fat (glyceride of one acid) so that we can continue with the calculation.