Question

Вычислите объем кислорода, необходимый для сжигания 500 м3 етину (объемы газов измерения при нормальных условиях) Помогите!

Answer 1

Формула этина C2H2 
2 C2H2 + 5 O2 = 4 CO2 + 2 H2O 
500 м³ = 500000 л
Т.к. вещества газы и условия нормальные, то соотношение коэффициентов является соотношением объемов. Тогда, 
V (O2) = (5 * 500000)/2 = 1250000 л = 1250 м³

Answer 2

Calculating the Volume of Oxygen Needed to Burn 500 m3 of Ethyne

To calculate the volume of oxygen needed to burn 500 m3 of ethyne (acetylene) at standard conditions, we can use the balanced chemical equation for the combustion of ethyne.

The balanced chemical equation for the combustion of ethyne is:

2C2H2 + 5O2 → 4CO2 + 2H2O

From the balanced equation, we can see that 2 moles of ethyne react with 5 moles of oxygen to produce 4 moles of carbon dioxide and 2 moles of water.

Given that the volume of gases is measured at standard conditions (STP), which is 1 mole of gas occupies 22.4 liters at 0°C and 1 atm, we can use the molar volume of gases to calculate the volume of oxygen needed.

Calculation:

1. Calculate the moles of ethyne: - Moles of C2H2 = Volume of C2H2 / Molar volume at STP - Moles of C2H2 = 500 m3 / 22.4 m3/mol

2. Use the mole ratio from the balanced equation to find moles of O2: - Moles of O2 = Moles of C2H2 * (5/2) - Moles of O2 = (500 m3 / 22.4 m3/mol) * (5/2)

3. Convert moles of O2 to volume at STP: - Volume of O2 = Moles of O2 * Molar volume at STP - Volume of O2 = (Moles of O2) * 22.4 m3/mol

Result:

The volume of oxygen needed to burn 500 m3 of ethyne at standard conditions is calculated to be the result of the third step.

Let's calculate the result using the given information.