Question

Для каталитического восстановления 5,34 г смеси нитробензола и n-нитротолуола потребовалось 2,688л водорода (н.у). Определите массовые доли веществ в исходной смеси.

Answer 1

Решение:
n(H2)=2,688/22,4=0,12моль
M(C6H5NO2)=123г/моль
M(C6H4CH3NO2)=137г/моль

хмоль--------3хмоль
C6H5NO2+3H2=======>C6H5NH2+2H2O
1моль-------3моль

(0,04-х) моль-----(0,12-3х) моль
C6H4CH3NO2+3H2==========>C6H4CH3NH2+2H2O
1моль--------------3моль

По условию задачи:
123х+137*(0,04-х) =5.34
х=0,01 моль
m(C6H5NO2)=123*0,01=1,23(г)
w(C6H5NO2)=1,23/5,34=0,23 или 23% это ответ.

Answer 2

Calculation of Mass Fractions in the Mixture

To determine the mass fractions of nitrobenzene and n-nitrotoluene in the given mixture, we can use the stoichiometry of the reaction between the mixture and hydrogen gas.

The balanced equation for the reaction is as follows:

C6H5NO2 + 3H2 → C6H12N2O2

According to the stoichiometry of the reaction, 1 mole of nitrobenzene reacts with 3 moles of hydrogen gas to produce 1 mole of the product, C6H12N2O2.

We are given that 5.34 g of the mixture reacts with 2.688 L of hydrogen gas at standard temperature and pressure (n.u). To determine the mass fractions, we need to calculate the moles of nitrobenzene and n-nitrotoluene in the mixture.

Let's assume the mass of nitrobenzene in the mixture is m1, and the mass of n-nitrotoluene is m2.

Using the molar masses of nitrobenzene (C6H5NO2) and n-nitrotoluene (C7H7NO2), which are approximately 123.11 g/mol and 137.14 g/mol respectively, we can calculate the moles of each substance.

The moles of nitrobenzene (n1) can be calculated using the formula:

n1 = m1 / M1

where m1 is the mass of nitrobenzene and M1 is the molar mass of nitrobenzene.

Similarly, the moles of n-nitrotoluene (n2) can be calculated using the formula:

n2 = m2 / M2

where m2 is the mass of n-nitrotoluene and M2 is the molar mass of n-nitrotoluene.

Since the total mass of the mixture is given as 5.34 g, we can write:

m1 + m2 = 5.34 g

Now, let's substitute the values and solve for the moles of nitrobenzene and n-nitrotoluene.

Using the given molar masses of nitrobenzene and n-nitrotoluene, we have:

n1 = m1 / 123.11

n2 = m2 / 137.14

Since the reaction stoichiometry tells us that 1 mole of nitrobenzene reacts with 3 moles of hydrogen gas, we can write:

n1 / 1 = n2 / 3

Simplifying this equation, we get:

n1 = (n2 / 3)

Substituting the expressions for n1 and n2, we have:

(m1 / 123.11) = (m2 / 3) / 137.14

Simplifying further, we get:

m1 = (m2 / 3) * (123.11 / 137.14)

Now, let's substitute the given values and calculate the mass fractions.

Given: - Volume of hydrogen gas (n.u) = 2.688 L - Mass of the mixture = 5.34 g

Using the ideal gas law, we can calculate the number of moles of hydrogen gas:

n(H2) = (P * V) / (R * T)

where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature.

At standard temperature and pressure (n.u), the pressure is 1 atm and the temperature is 273.15 K.

Substituting the values, we have:

n(H2) = (1 * 2.688) / (0.0821 * 273.15)

Now, let's calculate the moles of nitrobenzene and n-nitrotoluene using the equation:

n1 = (n2 / 3)

Finally, we can calculate the mass fractions using the formulas:

Mass fraction of nitrobenzene = (m1 / (m1 + m2)) * 100%

Mass fraction of n-nitrotoluene = (m2 / (m1 + m2)) * 100%

Let's calculate the mass fractions.