Answer to Question #201566 in Chemistry for Emma LoPresti

Question

Ethyne gas combusts with oxygen gas according to the following reaction:

Calculate the volume, in mL of CO2 produced when  84.4 g of C2H2 react at 62.2 °C and 1.38 atm. (R = 0.0821 L atm/mol K)

2𝐶2𝐻2(𝑔)+5𝑂2(𝑔)⟶4𝐶𝑂2(𝑔)+𝐻2𝑂(𝑙)

(round to hundredth place)

 

Answer 1

Solution:

Balanced chemical equation:

2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(l)

According to the equation above: n(C₂H₂) = n(CO₂)/2

The molar mass of C₂H₂ is 26.04 g/mol.

Therefore,

(84.4 g C₂H₂) × (1 mol C₂H₂ / 26.04 g C₂H₂) = 3.24 mol C₂H₂

n(CO₂) = 2 × n(C₂H₂) = 2 × 3.24 mol = 6.48 mol

According to the ideal gas law: PV = nRT

P = 1.38 atm

V = unknown

n = 6.48 mol

R = 0.0821 L atm mol^-1 K^-1

T = 62.2°C + 273.15 = 335.35 K

Therefore,

V = nRT/P

V(CO₂) = (6.48 mol × 0.0821 L atm mol^-1 K^-1 × 335.35 K) / (1.38 atm) = 129.28 L = 129281.80 mL

V(CO₂) = 129281.80 mL

Answer: 129281.80 mL of CO₂ produced.