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?
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.
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}