Question

При добавлении избытка натрия к 21,82 г смеси метанола, этанола и глицерина выделилось 7,178 дм3 водорода. из такого же количества смеси, используя HNO3, можно получить 27,24 г вещества, являющегося основой динамита. определите массы веществ в исходной смеси

Answer 1

2ch3oh+2na=2ch3ona+h2
2c2h5oh+2na=2c2h5ona+h2
2c3h5(oh)3+6na=2c3h5(ona)3+3h2

C3h5(oh)3+3hno3=c3h5(no3)+3h2o
n(c3h5(no3))=27,24/227=0,12 моль
n(c3h5(oh)3)=0,12 моль
n(h2)=7,178/22,4=0,32 моль-образовалось в 3-х реакциях
n(h2)=0,12/(2)*3=0,18 моль-образовалось в реакции с глицерином
m(c3h5(oh)3)=0,12*92=11,04 г
m(ch3oh+c2h5oh)=21,82-11,04=10,78 г
n(h2)=0,32-0,18=0,14моль- в реакциях с метанолом и этанолом

Составим систему уравнений
Пусть n(h2)в р-ции с метанолом= х моль, n(h2) в реакции с этанолом=y моль
x+y=0,14
2*32x+2*46y=10,78
Решая систему получаем
x=0,075 моль, y=0,065 моль
n(h2)в р-ции с метанолом=0,075 моль
m(ch3oh)=0,075*2*32=4,8 г
n(h2)в р-ции с этанолом=0,065 моль
m(c2h5oh)=0,065*2*46=5,98 г
Ответ:m(c3h5(oh)3)=11,04 г, m(c2h5oh)=5,98 г, m(ch3oh)=4,8 г

Answer 2

Calculation of Substance Masses in the Original Mixture

To determine the masses of substances in the original mixture, we need to use the given information about the addition of excess sodium and the volume of hydrogen gas produced. However, the search results provided do not contain any relevant information or equations to directly solve this problem. Therefore, I will guide you through the steps to solve it.

Let's assume the original mixture contains x grams of methanol, y grams of ethanol, and z grams of glycerin.

1. From the addition of excess sodium, we know that 7.178 dm³ of hydrogen gas was produced. We can use the ideal gas law to calculate the number of moles of hydrogen gas produced: - The volume of hydrogen gas can be converted to liters: 7.178 dm³ = 7.178 L. - The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. - Assuming the experiment was conducted at standard temperature and pressure (STP), we can use the values: P = 1 atm and T = 273.15 K. - The ideal gas constant R is 0.0821 L·atm/(mol·K). - Rearranging the ideal gas law equation to solve for n, we have n = PV / RT. - Substituting the values, we get n = (1 atm * 7.178 L) / (0.0821 L·atm/(mol·K) * 273.15 K). - Calculate the value of n.

2. The balanced equation for the reaction of sodium with water to produce hydrogen gas is: 2Na + 2H₂O → 2NaOH + H₂. From the equation, we can see that 2 moles of sodium react to produce 1 mole of hydrogen gas. - Use the stoichiometry of the balanced equation to determine the number of moles of sodium reacted. - Calculate the number of moles of sodium reacted.

3. Since we know the number of moles of sodium reacted, we can calculate the number of moles of each substance in the original mixture. - The number of moles of methanol is given by the ratio of the coefficient of methanol in the balanced equation to the coefficient of sodium: (1 mol methanol / 2 mol sodium) * moles of sodium reacted. - Similarly, calculate the number of moles of ethanol and glycerin.

4. Now, we can calculate the masses of each substance in the original mixture. - The mass of methanol is given by the number of moles of methanol multiplied by its molar mass. - Similarly, calculate the masses of ethanol and glycerin.

By following these steps, you should be able to determine the masses of the substances in the original mixture.