Consider the reaction: heat released

Question

Consider the reaction

B₂H₆(g) + 3 O₂(g) → B₂O₃(s) + 3 H₂O(g) ∆H = -2035 kJ

How much heat is released when a mixture of 4.37 g B₂H₆ and 3.81 g O₂ is burned?

heat released

Answer 1

B₂H₆(g) + 3 O₂(g) → B₂O₃(s) + 3 H₂O(g) ∆H = -2035 kJ

Obtain the molar quantities of B₂H₆(g) and O₂(g):

4.37g B₂H₆/27.67 g/mol = 0.158 mol B₂H₆

3.81 g O₂/16.00 g/mol = 0.238 mol O₂

Based on the coefficients of the reaction, O₂ is the limiting reagent because 0.238/3 = 0.079 mol O₂, which is less than the 0.158 mol B₂H₆.

Thus, 0.079 mol B₂H₆ and 0.238 mol O₂ will be fully consumed.

-2038 kJ is released when 1 mol of B₂H₆ reacts with 3 mol O₂.

Thus the amount of heat released: -2038 kJ/mol B₂H₆ * 0.079 mol = -161 kJ

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