Question

14. What volume of water vapor in liters could be generated at 1.01 atm and 420. K by the combustion of 273.36 grams of oxygen gas excess pentane gas (C5H12)?
C5H12(g) + 8 O2(g) ---> 5 CO2(g) + 6 H2O(g)

(OR C5 H12 ("g") + 8 O2 ("g") right arrow 5 C O2 ("g") + 6 H2 O ("g")




Do not type units with your answer

Answer 1

Answer:

776 L H₂O

Explanation:

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 vapor, 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.

(Step 1 & Step 2)

Molar Mass (C₅H₁₂): 5(12.011 g/mol) + 12(1.008 g/mol)

Molar Mass (C₅H₁₂): 72.151 g/mol

1 C₅H₁₂ (g) + 8 O₂ (g) -----> 5 CO₂ (g) + 6 H₂O (g)

273.36 g C₅H₁₂           1 mole            6 moles H₂O
-----------------------  x  ----------------  x  -----------------------  =  22.7 moles H₂O
                                  72.151 g           1 mole C₅H₁₂

(Step 3)

P = 1.01 atm                R = 0.0821 L*atm/mol*K

V = ?                           T = 420 K

n = 22.7 moles

PV = nRT

(1.01 atm)V = (22.7 moles)(0.0821 L*atm/mol*K)(420 K)

(1.01 atm)V = 783.856

V = 776 L

Without units, your answer would be 776.