Calculate the change in enthalpy when 52.0 g of solid chromium at 25°C and 1 atm pressure is oxidized. (H°f for Cr2O3(s) is –1135 kJ/mol.)

4Cr(s) + 3O2(g)  2Cr2O3(s)

a. –1135 kJ
b. –284 kJ
c. –568 kJ
d. +1135 kJ
e. +568 kJ

1 answer

The dH rxn = -1135 kJ/mol x 2 mol = ?
That is for 4*52 g Cr.
-1135*2 x (52/4*52) = ?