Answer
See answer below.
Work Step by Step
a) $P_4(s)+5\ O_2(g)\rightarrow P_4O_{10}(s)$
Enthalpy of formation:
$-2984\ kJ/mol$
Number of moles of white phosphorus:
$1.0\ g\div 123.89\ g/mol=8.07\ mmol$
Enthalpy change:
$-2984\ kJ/mol\times 8.07\ mmol=-24.09\ kJ$
b) $NO(g)\rightarrow 1/2\ N_2(g)+1/2\ O_2(g)$
Enthalpy of formation:
$+90.29\ kJ/mol$
Enthalpy change:
$-90.29\ kJ/mol\times 0.20\ mol=-18.06\ kJ$
c) $Na(s)+1/2\ Cl_2(g)\rightarrow NaCl(s)$
Enthalpy of formation:
$-411.12\ kJ/mol$
Number of moles of ammonia:
$2.40\ g\div 58.44\ g/mol=0.0411\ mol$
Enthalpy change:
$-411.12\ kJ/mol\times 0.0411\ mol=-16.88\ kJ$
d) $2\ Fe(s)+3/2\ O_2(g)\rightarrow Fe_2O_3(s)$
Enthalpy of formation:
$-825.3\ kJ/mol$
Number of moles of iron:
$250\ g\div 55.845\ g/mol=4.48\ mol$
Enthalpy change:
$-825.3\ kJ/mol\times 4.48\ mol\times1/2=-1847.3\ kJ$
