Answer
See answer below.
Work Step by Step
a) Enthalpies of formation (kJ/mol):
$NH_3(g): -45.90, NO(g): 90.29 , H_2O(g): -241.83$
Enthalpy of reaction:
$6\times-241.83+4\times90.29-4\times -45.90=-906.22\ kJ/mol$
b) Number of moles of ammonia:
$10.0\ g \div 17.03\ g/mol=0.587\ mol$
Enthalpy:
$-906.22\ kJ/mol\times0.587\ mol\times 1/4=-133.03\ kJ$
