Question

СРОЧНО!!! (задачі за хімічним рівнянням)1) Визначте об'єм кисню необхідний на спалювання етену об'ємом 20 л2) Визначити об'єм вуглекислого газу, що утворюється при згоранні бутану 11,2л​

Answer 1

Объяснение:

Начебто так, думаю що правильно.

Answer 2

Calculation of Oxygen Volume Required for Combustion of Ethene (C2H4)

To determine the volume of oxygen required for the combustion of ethene (C2H4), we need to consider the balanced chemical equation for the combustion reaction. The balanced equation for the combustion of ethene is as follows:

C2H4 + 3O2 → 2CO2 + 2H2O

From the balanced equation, we can see that 1 mole of ethene (C2H4) reacts with 3 moles of oxygen (O2) to produce 2 moles of carbon dioxide (CO2) and 2 moles of water (H2O).

To calculate the volume of oxygen required, we need to convert the given volume of ethene (20 L) to moles using the ideal gas law equation:

PV = nRT

Where: - P is the pressure of the gas (assumed constant) - V is the volume of the gas (20 L) - n is the number of moles of the gas (to be determined) - R is the ideal gas constant (0.0821 L·atm/mol·K) - T is the temperature of the gas (assumed constant)

Since the pressure and temperature are assumed constant, we can rearrange the equation to solve for the number of moles:

n = PV / RT

Now, we can calculate the number of moles of ethene (C2H4) using the given volume of 20 L:

n(C2H4) = (20 L) / (0.0821 L·atm/mol·K) n(C2H4) ≈ 243.3 mol

According to the balanced equation, 1 mole of ethene reacts with 3 moles of oxygen. Therefore, the number of moles of oxygen required can be calculated as:

n(O2) = 3 * n(C2H4) n(O2) ≈ 3 * 243.3 mol n(O2) ≈ 729.9 mol

Finally, we can convert the number of moles of oxygen to volume using the ideal gas law equation:

V(O2) = n(O2) * RT / P

Substituting the values, we get:

V(O2) = (729.9 mol) * (0.0821 L·atm/mol·K) / P

Since the pressure is not given in the question, we cannot calculate the exact volume of oxygen required without knowing the pressure. The volume of oxygen required will depend on the pressure at which the reaction is taking place.

Calculation of Carbon Dioxide Volume Produced from Combustion of Butane (C4H10)

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

2C4H10 + 13O2 → 8CO2 + 10H2O

From the balanced equation, we can see that 2 moles of butane (C4H10) react with 13 moles of oxygen (O2) to produce 8 moles of carbon dioxide (CO2) and 10 moles of water (H2O).

To calculate the volume of carbon dioxide produced, we need to convert the given volume of butane (11.2 L) to moles using the ideal gas law equation:

PV = nRT

Where: - P is the pressure of the gas (assumed constant) - V is the volume of the gas (11.2 L) - n is the number of moles of the gas (to be determined) - R is the ideal gas constant (0.0821 L·atm/mol·K) - T is the temperature of the gas (assumed constant)

Since the pressure and temperature are assumed constant, we can rearrange the equation to solve for the number of moles:

n = PV / RT

Now, we can calculate the number of moles of butane (C4H10) using the given volume of 11.2 L:

n(C4H10) = (11.2 L) / (0.0821 L·atm/mol·K) n(C4H10) ≈ 136.4 mol

According to the balanced equation, 2 moles of butane react with 8 moles of carbon dioxide. Therefore, the number of moles of carbon dioxide produced can be calculated as:

n(CO2) = 8 * n(C4H10) n(CO2) ≈ 8 * 136.4 mol n(CO2) ≈ 1091.2 mol

Finally, we can convert the number of moles of carbon dioxide to volume using the ideal gas law equation:

V(CO2) = n(CO2) * RT / P

Substituting the values, we get:

V(CO2) = (1091.2 mol) * (0.0821 L·atm/mol·K) / P

Since the pressure is not given in the question, we cannot calculate the exact volume of carbon dioxide produced without knowing the pressure. The volume of carbon dioxide produced will depend on the pressure at which the reaction is taking place.