Question

Визначте об'єм вуглекислого газу (л), що утвориться при спалюванні бутану масою 250кг, що містить 13% незаємих домішок. Відповідь запишіть без одиниць вимірювання - наприклад 4500 ...ДОПОМОЖІТЬ....​

Answer 1

To determine the volume of carbon dioxide gas produced during the combustion of butane, we need to consider the balanced chemical equation for the combustion reaction. The balanced equation for the combustion of butane (C4H10) is:

2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O

From the equation, we can see that 2 moles of butane react with 13 moles of oxygen to produce 8 moles of carbon dioxide.

To find the volume of carbon dioxide gas, we need to convert the mass of butane to moles using its molar mass, and then use the mole ratio from the balanced equation to find the moles of carbon dioxide. Finally, we can convert the moles of carbon dioxide to volume using the ideal gas law.

Here are the steps to calculate the volume of carbon dioxide gas:

1. Calculate the moles of butane: - Mass of butane = 250 kg - Molar mass of butane (C4H10) = 58.12 g/mol - Moles of butane = (250,000 g) / (58.12 g/mol)

2. Calculate the moles of carbon dioxide: - Moles of butane * (8 moles CO2 / 2 moles butane)

3. Convert moles of carbon dioxide to volume: - Volume of carbon dioxide = Moles of carbon dioxide * Molar volume of gas at STP

The molar volume of gas at STP (standard temperature and pressure) is approximately 22.4 L/mol.

Let's calculate the volume of carbon dioxide gas:

Calculation:

1. Moles of butane: - Mass of butane = 250 kg = 250,000 g - Molar mass of butane (C4H10) = 58.12 g/mol - Moles of butane = 250,000 g / 58.12 g/mol = 4302.89 mol

2. Moles of carbon dioxide: - Moles of carbon dioxide = 4302.89 mol * (8 mol CO2 / 2 mol butane) = 17211.56 mol

3. Volume of carbon dioxide: - Volume of carbon dioxide = 17211.56 mol * 22.4 L/mol = 385,242.98 L

Therefore, the volume of carbon dioxide gas produced during the combustion of 250 kg of butane, which contains 13% impurities, is approximately 385,243 liters.