For the reaction: B2O3 + 3H2O 3O2 +B2H6 H=2035KJ
How much energy is needed to react excess B2O3 with 25.0 g of H2O
Answer:- Energy needed = 941.6 kJ
Solution:-
B2O3(s) + 3H2O(g) ---> B2H6(g) + 3O2(g) ΔH = 2035 kJ
This means 2035 kJ of energy is needed to react excess B2O3 with 25.0 g of H2O
Lets convert 25.0 g of H2O into moles
Number of moles = weight/molecular weight = 25/ 18.01 = 1.38812 moles
1 mole of B2O3 reacts with 3 mole of H2O
so that
0.46271 mole of B2O3 reacts with 1.38812 mole of H2O
Energy needed = 0.46271 * 2035
Energy needed = 941.6 kJ
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