Respuesta :
589L volume of carbon dioxide in litres could be generated at 0.96 atm and 362. K by the combustion of 228.85 grams of oxygen gas excess pentane gas ([tex]C_5H_{12}[/tex])
What is an ideal gas equation?
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
[tex]C_5H_{12}(g) + 8 O_2(g)[/tex] --->[tex]5 CO_2(g) + 6 H_2O(g)[/tex]
To find the volume, you need to use the Ideal Gas Law. The equation looks like this:
PV = nRT
In this equation,
P = pressure (atm)
V = volume (L)
n = number of moles
R = constant (0.0821 L*atm/mol*K)
T = temperature (K)
To find the volume of water vapour, you need to (1) convert grams C₅H₁₂ to moles C₅H₁₂ (via molar mass), then (2) convert moles C₅H₁₂ to moles H₂O (via mole-to-mole ratio from reaction coefficients), then (3) calculate volume H₂O (via Ideal Gas Law). The final answer should have 3 sig figs.
Molar Mass (C₅H₁₂) =72.151 g/mol
1 C₅H₁₂ (g) + 8 O₂ (g) -----> 5 CO₂ (g) + 6 H₂O (g)
228.85 g C₅H₁₂ 1 mole 6 moles H₂O
----------------------- x ---------------- x ----------------------- = 19.03 moles H₂O
72.151 g 1 mole C₅H₁₂
Moles of H₂O= 19.03
P =0.96 atm
R = 0.0821 L x atm/mol x K
V = ?
T = 362 K
n = 19.03 moles
PV = nRT
(0.96 atm)V = ( 19.03 moles)(0.0821 Latm/molK)(362K)
(0.96 atm)V = 565.575406
V = 589L
Hence, 589L volume of carbon dioxide in litres could be generated at 0.96 atm and 362. K by the combustion of 228.85 grams of oxygen gas excess pentane gas ([tex]C_5H_{12}[/tex]).
Learn more about the ideal gas here:
https://brainly.com/question/27691721
#SPJ1
