Question

Спалили 115 г етанолу. Який об’єм кисню затрачен

Answer 1

Ответ:

168 л

Объяснение:

Смотри фото

Answer 2

Calculating the Volume of Oxygen Consumed in the Combustion of 115g of Ethanol

To calculate the volume of oxygen consumed in the combustion of 115g of ethanol, we can use stoichiometry to determine the balanced chemical equation for the combustion of ethanol and then use the molar ratio to find the volume of oxygen consumed.

Balanced Chemical Equation for the Combustion of Ethanol

The balanced chemical equation for the combustion of ethanol (C2H5OH) can be represented as:

C2H5OH + 3O2 → 2CO2 + 3H2O

This equation shows that one mole of ethanol reacts with three moles of oxygen to produce two moles of carbon dioxide and three moles of water.

Calculating the Molar Mass of Ethanol

The molar mass of ethanol (C2H5OH) can be calculated by adding the atomic masses of its constituent elements. According to the periodic table, the atomic masses are approximately: - Carbon (C): 12.01 g/mol - Hydrogen (H): 1.008 g/mol - Oxygen (O): 16.00 g/mol

Using these values, the molar mass of ethanol is: 12.01 * 2 + 1.008 * 6 + 16.00 + 1.008 = 46.07 g/mol

Calculating the Moles of Ethanol

To find the moles of ethanol in 115g, we can use the formula: moles = mass / molar mass

So, the moles of ethanol = 115g / 46.07 g/mol = 2.496 moles

Using the Stoichiometric Coefficients to Find the Moles of Oxygen

From the balanced chemical equation, we know that 1 mole of ethanol reacts with 3 moles of oxygen. Therefore, the moles of oxygen required can be calculated as: moles of oxygen = moles of ethanol * (3 moles of O2 / 1 mole of C2H5OH)

Substituting the value of moles of ethanol: moles of oxygen = 2.496 moles * 3 = 7.488 moles

Calculating the Volume of Oxygen Consumed

To find the volume of oxygen consumed, we can use the ideal gas law, which states that: PV = nRT

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

Assuming standard temperature and pressure (STP), where T = 273 K and P = 1 atm, we can rearrange the ideal gas law to solve for V: V = nRT / P

Substituting the values: V = 7.488 moles * 0.0821 L·atm/mol·K * 273 K / 1 atm ≈ 168.7 L

Therefore, the volume of oxygen consumed in the combustion of 115g of ethanol at STP is approximately 168.7 liters.