Question

What 8s the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

Answer 1

The chemical equation for the combustion of ethylene is as follows:

C2H4 (ethylene) +3O2———›2CO2+2H2O

Number of moles = amount in g / molar mass

Given that 2.8 kg ethylene = 2800 g  

Number of moles = 2800 g/ 28.05 g/ mole = 99.82 mole

Moles of O2 = 99.82 mole C2H4 * 3  mole O2 / 1 mole C2H4

= 299.46 mole O2

Amount of O2 = number of moles * molar mass

=  299.46 mole O2 *32 g/ mole

= 9582.72 g

= 9.6 kg

Answer 2

Answer:

The chemical equation for the said reaction is,

C2H4+12×2+4=283O296→2CO2+2H2O

the gram molar weight of ethylene is= 12*2+1*4 g

=28g

the gram molar weight of oxygen = 3*32 g

=96 g

∵ The weight of oxygen required for complete combustion of 28 g ethylene =96 g

according to question,

∴ Weight of oxygen required for combustion of 2.8 kg ethylene =96×2.8×100028×1000kg

=9.6kg

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