First slide

Enthalpy of Combustion

Question

When one mole of manganese is burnt completely to form Mn₃O_4(s) at standard state conditions, the amount of heat liberated is
$(\Delta H_f^0\;of\;M{n_3}{O_4} = {\rm{ - }}1388kJmo{l^{ - 1}})$

Easy

Solution

$3 Mn (s) + 2 O_{2} (g) \to {Mn}_{3} O_{4} (s),$ ΔH=-1388KJ/mole
For 3 mole Mn heat liberated is 1388KJ
1 mole Mn -------?
$= \;\frac{{1388}}{3}\; = \;462.67KJ\$
Heat liberated = 462.67KJ

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App