Question

Помогите пожалуйста!! объем бутана при нормальных условиях, при сжигании которого выделилось 134,4 литра углекислого газа

Answer 1

2С4H10+13O2=8CO2+10H2O
V(C4H10)=134,4л*44,8л/179,2л=33,6л

Answer 2

Calculating the Volume of Butane Gas

To calculate the volume of butane gas at standard conditions (0 degrees Celsius and 1 atmosphere) that produced 134.4 liters of carbon dioxide when burned, we can use the balanced chemical equation for the combustion of butane and the ideal gas law.

The balanced chemical equation for the combustion of butane is: 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

According to the stoichiometry of the reaction, 2 moles of butane produce 8 moles of carbon dioxide. Therefore, we can calculate the volume of butane gas using the ideal gas law equation:

PV = nRT

Where: - P = pressure (atm) - V = volume (L) - n = number of moles - R = ideal gas constant (0.0821 L·atm/mol·K) - T = temperature (K)

Calculation

Given that 134.4 liters of carbon dioxide were produced, we can calculate the number of moles of carbon dioxide produced using the balanced chemical equation and then use this information to find the volume of butane gas.

1. Calculate the moles of carbon dioxide produced: - 134.4 liters of CO2 at standard conditions can be converted to moles using the ideal gas law and the molar volume of a gas at standard conditions (22.4 L/mol). - Moles of CO2 = Volume of CO2 / Molar volume of gas at standard conditions

2. Use the stoichiometry of the reaction to find the moles of butane: - Since 2 moles of butane produce 8 moles of CO2, we can calculate the moles of butane consumed in the reaction.

3. Calculate the volume of butane gas: - Once the moles of butane are known, we can use the ideal gas law to calculate the volume of butane gas at standard conditions.

Conclusion

By following these steps, you can calculate the volume of butane gas at standard conditions that produced 134.4 liters of carbon dioxide when burned.