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)

Expert's answer

**Solution:**

Balanced chemical equation:

2C_{2}H_{2}(g) + 5O_{2}(g) → 4CO_{2}(g) + 2H_{2}O(l)

According to the equation above: n(C_{2}H_{2}) = n(CO_{2})/2

The molar mass of C_{2}H_{2} is 26.04 g/mol.

Therefore,

(84.4 g C_{2}H_{2}) × (1 mol C_{2}H_{2} / 26.04 g C_{2}H_{2}) = 3.24 mol C_{2}H_{2}

n(CO_{2}) = 2 × n(C_{2}H_{2}) = 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(CO_{2}) = (6.48 mol × 0.0821 L atm mol^{-1} K^{-1} × 335.35 K) / (1.38 atm) = 129.28 L = 129281.80 mL

V(CO_{2}) = 129281.80 mL

**Answer: 129281.80 mL of CO**_{2}** produced.**

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