Question

Який об’єм (л) карбон (ІV) оксиду утвориться під час спалювання 224 л етину?помогите пж

Answer 1

Ответ:

С2Н4 +3О2 = 2СО2 +2Н2О

Молярная масса этилена С2Н4 равна 12*2+4*1=28г

Количество вещества этилена С2Н4 равно 2,8/28=0,1моль

Количество вещества СО2 равно 0,1*2=0,2моль

Объем СО2 равен 0,2*22,4 = 4,48л

Answer 2

Calculating the Volume of Carbon(IV) Oxide Produced

To calculate the volume of carbon(IV) oxide (CO2) produced during the combustion of 224 liters of ethyne (also known as acetylene), 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(IV) oxide.

Calculation

Given that the volume of ethyne is 224 liters, we can use the mole ratio from the balanced equation to calculate the volume of carbon(IV) oxide produced.

1. Calculate the moles of ethyne: - Moles of ethyne = Volume of ethyne / molar volume of ethyne - Moles of ethyne = 224 L / 22.4 L/mol (molar volume of any gas at STP) - Moles of ethyne = 10 moles

2. Use the mole ratio to find the moles of carbon(IV) oxide produced: - Moles of CO2 = Moles of ethyne × (4 moles of CO2 / 2 moles of ethyne) - Moles of CO2 = 10 moles × 2 - Moles of CO2 = 20 moles

3. Convert moles of carbon(IV) oxide to volume: - Volume of CO2 = Moles of CO2 × molar volume of CO2 - Volume of CO2 = 20 moles × 22.4 L/mol (molar volume of any gas at STP) - Volume of CO2 = 448 liters

Conclusion

During the combustion of 224 liters of ethyne, 448 liters of carbon(IV) oxide will be produced.