How many moles of ethylene (C2H4) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin? C2H4 (g) + 3O2 (g) yields 2CO2 (g) + 2H2O (g) 0.21 moles 0.32 moles 0.63 moles 0.84 moles

Question

Adriana Mcintosh

Chemistry

9 August, 12:54

How many moles of ethylene (C2H4) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin? C2H4 (g) + 3O2 (g) yields 2CO2 (g) + 2H2O (g) 0.21 moles 0.32 moles 0.63 moles 0.84 moles

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Haylie Cisneros · Answer 1

1) Convert 12.9 liters of Oxygen to mol at the given conditions:

PV = nRT ⇒ n = PV/RT

n = [1.2atm*12.9 l] / [0.082 atm l / K mol * 297K]

n = 0.636 mol of O2

2) use the stoichiometry derived from the balanced chemical equation

1mol C2H4 / 3 mol O2 = x mol C2H4 / 0.636 mol O2

x = 0.636 / 3 mol O2 = 0.212 mol O2.

Answer: 0.212 mol O2