Respuesta :
the balanced equation for the reaction is as follows
2C₄H₁₀ + 13O₂ ---> 8 CO₂ + 10H₂O
stoichiometry of C₄H₁₀ to O₂ is 2:13
stoichiometry applies to the molar ratio of reactants and products. Avagadros law states that volume of gas is directly proportional to number of moles of gas when pressure and temperature are constant.
Therefore volume ratio of reactants is equal to molar ratio, volume ratio of C₄H₁₀ to O₂ is 2:13
2 L of C₄H₁₀ reacts with 13 L of O₂
then 100 L of C₄H₁₀ reacts with 13/2 x 100 = 650 L
therefore 650 L of O₂ are required
2C₄H₁₀ + 13O₂ ---> 8 CO₂ + 10H₂O
stoichiometry of C₄H₁₀ to O₂ is 2:13
stoichiometry applies to the molar ratio of reactants and products. Avagadros law states that volume of gas is directly proportional to number of moles of gas when pressure and temperature are constant.
Therefore volume ratio of reactants is equal to molar ratio, volume ratio of C₄H₁₀ to O₂ is 2:13
2 L of C₄H₁₀ reacts with 13 L of O₂
then 100 L of C₄H₁₀ reacts with 13/2 x 100 = 650 L
therefore 650 L of O₂ are required
Answer:
Volume of O2 = 649 L
Explanation:
Given data:
Volume of butane, V = 100 L
Temperature , T = 25 C = 25 + 273 = 298 K
Standard Pressure, P = 1 atm
Step 1: Calculate the moles of butane using ideal gas equation
[tex]PV = nRT[/tex]
where n = number of moles, R = 0.0821 Latm/mol-K
[tex]n = \frac{PV}{RT} = \frac{1*100}{0.0821*298} = 4.087 moles[/tex]
Step 2: Calculate moles of O2 required from Reaction stoichiometry
2 C₄H₁₀(g) + 13 O₂(g) ⇄ 8 CO₂(g) + 10H₂O(l)
Based on the stoichiometry:
2 moles of butane reacts with 13 moles of O2
Therefore, 4.087 moles of butane will react with:
[tex]\frac{4.078 butane*13 O2}{2 butane } = 26.507 moles O2[/tex]
Step 3: Calculate the volume of O2 required
Again from ideal gas equation:
[tex]V = \frac{nRT}{P} = \frac{26.507*0.0821*298}{1} = 648.5 L[/tex]
