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7.In the oxidation of ethane: 2 C2H6 + 7 02 + 4CO2 + 6H2O how many
liters of O2 are required to react with 90 grams of ethane?


Sagot :

Answer:

If reaction is taking place at room temperature and pressure(rtp)

= 360 litres of O₂ at rtp

If reaction is taking place at  standard temperature and pressure(stp):

= 336 litres of O₂ at stp

Explanation:

2C₂H₆ + 70₂ → 4CO₂ + 6H₂O

1 mole of C₂H₆ =  (12 × 2) + (1 × 6) = 24 + 6 = 30g

2 moles of C₂H₆ = 30 × 2 = 60g

From the equation:

1 mole of 0₂ reacts with 2 moles of C₂H₆

1 mole of 0₂ reacts with  60g  of C₂H₆

? moles of 0₂ react with 90g of C₂H₆

[tex]\frac{90}{60}[/tex]= 1.5 moles of O₂

Considering condition of reaction taking place whether room temperature and pressure(rtp) or standard temperature and pressure(stp)

If room temperature and pressure(rtp):

1 mole of 0₂ occupies 24 liters/24000 cm³/ 24 dm³ of O₂

1.5 moles of O₂ occupies (24 × 1.5)  litres of O₂

= 360 litres of O₂ at rtp

If standard temperature and pressure(stp):

1 mole of 0₂ occupies 22.4 liters/22400 cm³/ 22.4 dm³ of O₂

1.5 moles of O₂ occupies (22,4 × 1.5)  litres of O₂

= 336 litres of O₂ at stp