Question

Зазнач об'єм вуглекислого газу, який утвориться в результаті згоряння 527,8 г бутану. Буду очень рад отведу, спасибо

Answer 1

Ответ:

815,36 л С4Н10

Объяснение:

фото

Answer 2

Calculating the Volume of Carbon Dioxide Produced from the Combustion of Butane

To calculate the volume of carbon dioxide produced from the combustion of 527.8 grams of butane, we can use the balanced chemical equation for the combustion of butane and the ideal gas law.

The balanced chemical equation for the combustion of butane is:

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

This equation shows that 2 moles of butane react with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.

First, we need to convert the mass of butane to moles using its molar mass.

Molar Mass of Butane

The molar mass of butane (C4H10) can be calculated as follows: - Carbon (C) has a molar mass of approximately 12.01 g/mol. - Hydrogen (H) has a molar mass of approximately 1.01 g/mol.

So, the molar mass of butane (C4H10) is: 4 * 12.01 (for carbon) + 10 * 1.01 (for hydrogen) = 58.12 g/mol

Now, we can calculate the number of moles of butane.

Moles of Butane

Moles of butane = Mass of butane / Molar mass of butane Moles of butane = 527.8 g / 58.12 g/mol Moles of butane ≈ 9.08 moles

Now that we have the moles of butane, we can use the balanced chemical equation to determine the moles of carbon dioxide produced.

Moles of Carbon Dioxide Produced

From the balanced chemical equation, we know that 2 moles of butane produce 8 moles of carbon dioxide. So, moles of carbon dioxide = (moles of butane) * (8 moles of CO2 / 2 moles of C4H10) Moles of carbon dioxide = 9.08 moles * (8 / 2) Moles of carbon dioxide = 36.32 moles

Now, we can use the ideal gas law to calculate the volume of carbon dioxide produced.

Ideal Gas Law

The ideal gas law is given by the equation: PV = nRT

Where: P = pressure V = volume n = number of moles R = ideal gas constant T = temperature

Assuming standard temperature and pressure (STP), which is 0°C (273.15 K) and 1 atmosphere (1.00 atm), the ideal gas constant (R) is approximately 0.0821 L·atm/(K·mol).

Calculating Volume of Carbon Dioxide

V = (nRT) / P V = (36.32 moles * 0.0821 L·atm/(K·mol) * 273.15 K) / 1.00 atm V ≈ 808.7 L

The volume of carbon dioxide produced from the combustion of 527.8 grams of butane at STP is approximately 808.7 liters.