Answer to Question #201566 in Chemistry for Emma LoPresti

Question #201566

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)

 


1
Expert's answer
2021-06-02T06:02:59-0400

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.

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