Question

You are given the following problem to solve. The ideal gas constant for this equation will be

0.0821 L(atm)/mol(K).
How many liters of CO, will be produced when 15.8 g of butane (CH,o) are burned at 128°C
at 0.98 atm?

Answer 1

V = 9.14 L

Step-by-step explanation:

To find the volume of butane, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = number of moles

-----> R = Ideal Gas Constant (L*atm/mol*K)

-----> T = temperature (K)

Before you can solve for the volume, you need to (1) convert the mass of butane to moles (using the molar mass) and then (2) convert the temperature from Celsius to Kelvin.

Atomic Mass (C): 12.011 g/mol

Atomic Mass (H): 1.008 g/mol

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol) = 58.124 g/mol

15.8 g C₄H₁₀ 1 mol
--------------------- x ----------------- = 0.272 moles C₄H₁₀
58.124 g

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P = 0.98 atm R = 0.0821 L*atm/mol*K

V = ? L T = 128°C + 273 = 401 K

n = 0.272 moles

PV = nRT <----- Ideal Gas Law

(0.98 atm)V = (0.272 moles)(0.0821 L*atm/mol*K)(401 K) <----- Insert variables

(0.98 atm)V = 8.9548 L*atm <----- Multiply right side

V = 9.14 L <----- Divide both sides by 0.98