Question

Рассчитайте объём оксида углерода(IV) при сгорании 3.2 г метана (цэ-аш-четыре)?

Answer 1

СН4 + 2О2= СО2 + 2Н2О
n(CH4)= m/Mr
Mr( CH4)= 12+ 4= 16(г/моль)
n(CH4)= 3.2/16 = 0.2(моль)
n( CO2) = 0.2 моль
V(CO2) n*Vm
Vm = 22.4 дм3/ моль
V(CO2) = 0.2 * 22.4 = 4.48(дм3)
Ответ: 4.48 дм3

Answer 2

Calculating the Volume of Carbon Dioxide (IV) Produced from the Combustion of Methane

To calculate the volume of carbon dioxide (IV) produced from the combustion of 3.2 grams of methane, we can use the balanced chemical equation for the combustion of methane.

The balanced chemical equation for the combustion of methane is: CH4 + 2O2 → CO2 + 2H2O

From the balanced chemical equation, we can see that 1 mole of methane produces 1 mole of carbon dioxide (IV) when it combusts.

Using the molar mass of methane (CH4) and carbon dioxide (CO2), we can calculate the volume of carbon dioxide produced.

The molar mass of methane (CH4) is approximately 16 g/mol, and the molar mass of carbon dioxide (CO2) is approximately 44 g/mol.

Calculation

1. Calculate the moles of methane: - Moles of CH4 = Mass of CH4 / Molar mass of CH4 - Moles of CH4 = 3.2 g / 16 g/mol - Moles of CH4 = 0.2 moles

2. Since the balanced chemical equation shows that 1 mole of methane produces 1 mole of carbon dioxide, the moles of carbon dioxide produced will also be 0.2 moles.

3. Now, we can use the ideal gas law to calculate the volume of carbon dioxide produced at standard temperature and pressure (STP), which is 0°C and 1 atm pressure.

- The molar volume of a gas at STP is 22.4 L/mol.

- Volume of CO2 = Moles of CO2 * Molar volume at STP - Volume of CO2 = 0.2 moles * 22.4 L/mol - Volume of CO2 = 4.48 L

Therefore, the volume of carbon dioxide (IV) produced from the combustion of 3.2 grams of methane is approximately 4.48 liters.

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