Question

Зазнач об'єм вуглекислого газу, який утвориться в результаті згоряння 107,2 г метану.

Answer 1

Ответ:

6,7. ×

CH4+2O2-->CO2+2H2O

1 1

Объяснение:

n(CH4)=m(CH4):M(CH4)=107,2:16=6,7моль

х=6,7×1:1=6,7моль

V(CO2)=n(CO2)×Vm=150,08л

Ответ:158,08 литров

Answer 2

Volume of Carbon Dioxide Produced from the Combustion of 107.2g of Methane

When 107.2 grams of methane undergoes combustion, the volume of carbon dioxide produced can be calculated using 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 equation, it can be observed that 1 mole of methane (CH4) reacts with 2 moles of oxygen (O2) to produce 1 mole of carbon dioxide (CO2) and 2 moles of water (H2O).

To calculate the volume of carbon dioxide produced, we can use the ideal gas law, which states that at standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.

First, we need to convert the mass of methane to moles using its molar mass. The molar mass of methane (CH4) is approximately 16 g/mol.

Calculations: - Number of moles of methane = Mass of methane / Molar mass of methane - Number of moles of methane = 107.2 g / 16 g/mol - Number of moles of methane ≈ 6.7 moles

According to the balanced chemical equation, 1 mole of methane produces 1 mole of carbon dioxide. Therefore, the number of moles of carbon dioxide produced is also approximately 6.7 moles.

Finally, to find the volume of carbon dioxide at STP, we can use the relationship that 1 mole of any gas at STP occupies 22.4 liters.

Volume of carbon dioxide produced ≈ Number of moles of carbon dioxide * 22.4 liters/mole

Volume of carbon dioxide produced ≈ 6.7 moles * 22.4 liters/mole

Volume of carbon dioxide produced ≈ 150.08 liters

Therefore, approximately 150.08 liters of carbon dioxide will be produced from the combustion of 107.2 grams of methane.