Question

14. How many kilojoules of heat energy will be liberated when 49.70 g of manganese are burned to form Mn3O4 (s) at standard state conditions? Hof298 of Mn3O4 is equal to -1388 kJ mol-1.

Answer 1

Given that;

The standard state Ho f 298 of Mn3O4 is equal to -1388 kJ mol-1

Mn3O4(s) ---> 3Mn(g) + 2O2(g), ΔH = ?
-1,388 kJ/mole à 283.3 kJ/mole, 0 kJ/mole

First calculate the heat of the above reaction:
ΔH = Hproducts – Hreactants =
(283.3 + 0) – (-1,388) = 283.3 + 1,387.8 = 1,671.3 kJ/mole

Mn3O4(s) ---> 3Mn(g) + 2O2(g), ΔH = 1,671.3 kJ/mole

Now calculate the number of moles of Mn= amount in g/ molar mass

= 49.70 g/ 54.94g/mol

= 0.905 mol Mn

Now calculate the energy in kilojoules which will be liberated when 49.70 g of manganese are burned to form Mn3O4 (s) as follows:

1,671.3 kJ/mole * 0.905 mol =

1512.5 kJ