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2H\(_2\) + O\(_2\) → 2H\(_2\)O    ΔH = - 571kJ In the equation above, how...

Chemistry
JAMB 2005

2H\(_2\) + O\(_2\) → 2H\(_2\)O    ΔH = - 571kJ
In the equation above, how much heat will be liberated if 12.0g of hydrogen is burnt in excess oxygen?

  • A. -1142 kJ
  • B. -571 kJ
  • C. +1142 kJ
  • D. -1713 kJ
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Correct Answer: Option D
Explanation

2H\(_2\) + O\(_2\) → 2H\(_2\)O   ΔH = -571kJ

For Hydrogen, Number of moles = \(\frac{mass}{Molar mass}\)

Stoichiometrically,       2    =   \(\frac{mass}{2}\)

   Mass of Hydrogen = 2 x 2 = 4g
 If 4.0g of Hydrogen burns in O\(_2\) to produce - 571 kJ of energy
∴ 12.0g of Hydrogen will burn in O\(_2\) to produce  x kJ of energy
        x  =  \(\frac{{12}\times{- 571}}{4}\)

            =  - 1713kJ

There is an explanation video available below.


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