1. 5.76 4 B + 3 O2  2 B2O3 O2 + 2 H2  2 H2O 3O2 + B2H6  B2O3 + 3 H2O
THERE ARE ONLY 2 B IN THE NET, YOU MUST MULTIPLY THIS REACTION BY ½.
THERE ARE 3 H IN THE NET, YOU MUST MULTIPLY THIS REACTION BY 3/2.
4 B + 3 O2  2 B2O3
∆H1=( kJ)
O2 + 2 H2  2 H2O
∆H2=( kJ)
3O2 + B2H6  B2O3 + 3 H2O
∆H3= kJ)
REVERSE THIS REACTION TO GET B2H6 AS A REACTANT. AND TO CANCEL B2O3
2B + 3H2  B2H6
This is the formation, ∆Hf of B2H6, NET Rx

2. This is the formation, ∆Hf of B2H6, NET Rx
2 B + 3/2 O2  B2O3
∆H1=1/2( kJ)
3/2O2 + 3 H2  3 H2O
∆H2=3/2( kJ)
B2O3 + 3 H2O  B2H6 + 3O2
∆H3=-( kJ)
2B + 3H2  B2H6
This is the formation, ∆Hf of B2H6, NET Rx
∆Hf=+35.4 kJ