Question

Рассчитайте объём углекислого газа (н.у.), полученного при полном сгорании 60 г углерода? *Нужна помощь! Пожалуйста)*

Answer 1

Реакция полного сгорания углерода
C + O₂ = CO₂

по реакции из 1 моля углерода образуется 1 моль углекислого газа
n(CO₂)=n(C) (1)

количество вещества углерода
n(C)=m(C)/M(C)   (2)
m(C)=60 г масса углерода
M(C)=12 г/моль молярная масса углерода

объем углекислого газа
V(CO₂)=V₀*n(CO₂)   (3)
V₀=22,4 л/моль - молярный объем газа

подставляем в (3) из (2) и (1):
V(CO₂)=V₀*m(C)/M(C)
V(CO₂)=22,4л/моль*60г/12г/моль=112 л

Answer 2

Calculating the Volume of Carbon Dioxide Gas Produced

To calculate the volume of carbon dioxide gas produced from the complete combustion of 60 grams of carbon, we can use the stoichiometry of the reaction. The balanced chemical equation for the complete combustion of carbon is:

C(s) + O2(g) → CO2(g)

From the balanced equation, we can see that 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide.

Step 1: Convert Mass of Carbon to Moles

First, we need to convert the mass of carbon to moles. The molar mass of carbon is approximately 12 grams per mole.

Using the formula:

moles = mass / molar mass

We can calculate the moles of carbon:

moles = 60g / 12g/mol = 5 moles

Step 2: Calculate Moles of Carbon Dioxide Produced

Since the stoichiometry of the reaction is 1:1 between carbon and carbon dioxide, the moles of carbon dioxide produced will also be 5 moles.

Step 3: Convert Moles to Volume at Standard Conditions

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

Therefore, the volume of carbon dioxide gas produced will be:

volume = moles * 22.4 liters/mol

volume = 5 moles * 22.4 liters/mol = 112 liters

So, the volume of carbon dioxide gas produced from the complete combustion of 60 grams of carbon is 112 liters.

This calculation is based on the stoichiometry of the reaction and the principles of ideal gas behavior.

If you have any further questions or need additional assistance, feel free to ask!