Question

Помогите, пожалуйста!Определите объем азота (н.у.), который образуется при горении 31 грамма метиламина.Ответ запишите с точностью до десятых.Уравнение хим. реакции:4CH3NH2 + 9O2---》4CO2 + 10H2O + 2N2

Answer 1

n(CH3NH2) = m/M = 31г/31г/моль = 1моль

По уравнению реакции 2n(N2) = 4n(CH3NH2), n(N2) = 2n(CH3NH2)

n(N2) = 1/2 = 0,5моль

V(N2) = n x Vm = 0,5моль х 22,4л/моль = 11,2л

Answer 2

Calculating the Volume of Nitrogen Gas Produced

To calculate the volume of nitrogen gas (N2) produced when 31 grams of methylamine (CH3NH2) burns, we can use the given chemical equation and the concept of stoichiometry.

The balanced chemical equation for the combustion of methylamine is: 4CH3NH2 + 9O2 → 4CO2 + 10H2O + 2N2

Steps to Calculate the Volume of Nitrogen Gas Produced:

1. Convert Mass to Moles: Convert the given mass of methylamine to moles. 2. Use Stoichiometry: Use the stoichiometric coefficients from the balanced chemical equation to find the moles of nitrogen gas produced. 3. Convert Moles to Volume: Convert the moles of nitrogen gas to volume at standard conditions.

Let's proceed with the calculations.

Step 1: Convert Mass to Moles

The molar mass of methylamine (CH3NH2) can be calculated as follows: - Carbon (C): 12.01 g/mol - Hydrogen (H): 1.01 g/mol - Nitrogen (N): 14.01 g/mol

Molar mass of CH3NH2 = (1 * 12.01) + (5 * 1.01) + 14.01 = 31.06 g/mol

Now, we can calculate the moles of methylamine: Moles = Mass / Molar mass Moles = 31 g / 31.06 g/mol Moles ≈ 1 mol

Step 2: Use Stoichiometry

From the balanced chemical equation, we can see that 4 moles of methylamine produce 2 moles of nitrogen gas (N2).

Step 3: Convert Moles to Volume

At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.

So, the volume of nitrogen gas produced can be calculated as: Volume = Moles of N2 * 22.4 L/mol Volume = 1 mol * 22.4 L/mol Volume = 22.4 L

Answer:

The volume of nitrogen gas (N2) produced when 31 grams of methylamine burns is approximately 22.4 liters.