Question

Определите массу и объем азота (н.у), образовавшегося при сгорании 4,5г этиламина. Какой объем воздуха необходим для сжигания данной порции этиламина (объемная доля кислорода в воздухе 21%) ?

Answer 1

4C2H5NH2 + 15O2 = 8CO2 + 14H2O + 2N2
n = m/M = V/Vm
n(C2H5NH2) = 4,5/45 = 0,1 моль
n(N2) = 0,1*2/4 = 0,05 моль
V(N2) = 0,05*22,4 = 1,12 л
m(N2) = 0,05*28 = 1,4 г

n(O2) = 0,1*15/4 = 0,375 моль
V(O2) = 0,375*22,4 = 8,4 л,
т.к. кислорода в воздухе 21%:
V(возд) = 8,4/0,21 = 40 л

Answer 2

Calculating the Mass and Volume of Nitrogen (N2) Produced

To calculate the mass and volume of nitrogen (N2) formed during the combustion of 4.5g of ethylamine, we can use the balanced chemical equation for the combustion of ethylamine and the ideal gas law.

The balanced chemical equation for the combustion of ethylamine is: 2 C2H5NH2 + 9 O2 → 4 CO2 + 6 H2O + 2 N2

First, let's calculate the moles of ethylamine (C2H5NH2) and then use the stoichiometry of the balanced chemical equation to find the moles of nitrogen (N2) produced.

The molar mass of ethylamine (C2H5NH2) is 45.08 g/mol. The molar mass of nitrogen (N2) is 28.02 g/mol.

Step 1: Calculate the moles of ethylamine Moles of ethylamine = Mass of ethylamine / Molar mass of ethylamine Moles of ethylamine = 4.5g / 45.08 g/mol Moles of ethylamine = 0.0999 mol

Step 2: Use stoichiometry to find moles of nitrogen From the balanced chemical equation, 2 moles of ethylamine produce 2 moles of nitrogen. So, moles of nitrogen = Moles of ethylamine * (2 moles of N2 / 2 moles of C2H5NH2) Moles of nitrogen = 0.0999 mol * (2 / 2) Moles of nitrogen = 0.0999 mol

Step 3: Calculate the mass of nitrogen Mass of nitrogen = Moles of nitrogen * Molar mass of nitrogen Mass of nitrogen = 0.0999 mol * 28.02 g/mol Mass of nitrogen = 2.802 g

Step 4: Calculate the volume of nitrogen at standard conditions (0°C and 1 atm) Using the ideal gas law: PV = nRT, where P = pressure, V = volume, n = moles, R = ideal gas constant, and T = temperature.

At standard conditions, the molar volume of an ideal gas is 22.4 L/mol. So, volume of nitrogen = Moles of nitrogen * Molar volume at STP Volume of nitrogen = 0.0999 mol * 22.4 L/mol Volume of nitrogen = 2.239 L

Volume of Air Required for Combustion

To find the volume of air required for the combustion of the given amount of ethylamine, we need to consider the stoichiometry of the reaction and the volume percentage of oxygen in air.

The balanced chemical equation for the combustion of ethylamine is: 2 C2H5NH2 + 9 O2 → 4 CO2 + 6 H2O + 2 N2

From the balanced equation, 9 moles of oxygen are required to combust 2 moles of ethylamine.

Given that the volume percentage of oxygen in air is 21%, we can use this information to calculate the volume of air required.

Step 1: Calculate the moles of ethylamine (Already calculated in the previous section) Moles of ethylamine = 0.0999 mol

Step 2: Calculate the moles of oxygen required From the balanced chemical equation, 2 moles of ethylamine require 9 moles of oxygen. Moles of oxygen = Moles of ethylamine * (9 moles of O2 / 2 moles of C2H5NH2) Moles of oxygen = 0.0999 mol * (9 / 2) Moles of oxygen = 0.4496 mol

Step 3: Calculate the volume of air required Using the molar volume of an ideal gas at standard conditions (22.4 L/mol) and the volume percentage of oxygen in air (21%), we can calculate the volume of air required. Volume of air = Moles of oxygen * (100 / 21) * Molar volume at STP Volume of air = 0.4496 mol * (100 / 21) * 22.4 L/mol Volume of air = 9.57 L

Therefore, the volume of air required for the combustion of the given amount of ethylamine is approximately 9.57 liters.

I hope this helps! If you have further questions, feel free to ask.