Question

he equation below describes the reaction of diborane with oxygen gas.

B₂H₆(l) + 3 O₂(g) → B₂O₃(s) + 3 H₂O(l)

If 35.6 mL of B₂H₆ reacted with excess oxygen gas, determine the actual yield of B₂O₃ if the percent yield of B₂O₃ was 77.0%. (The density of B₂H₆ is 1.131 g/mL. The molar mass of B₂H₆ is 27.668 g/mol and the molar mass of B₂O₃ is 69.62 g/mol.)

Answer 1

B₂H₆(l) + 3 O₂(g) → B₂O₃(s) + 3 H₂O(l)

mass of B2H6   = volume * density

                          = 35.6*1.131   = 40.2636g

B₂H₆(l) + 3 O₂(g) → B₂O₃(s) + 3 H₂O(l)

1 mole of B2H6 react excess of oxygen to gives 1 moles of B2O3

22.669g of B2H6 react with excess of oxygen to gives 69.62g of B2O3

40.2636g of B2H6 react with excess of oxygen to gives = 69.62*40.2636/22.669    = 123.65g of B2O3

    theoretical yield of B2O3 = 123.65g

percent yiled   = actual yield*100/theoretical yiled

77                   = actual yiled*100/123.65

actual yield         = 77*123.65/100   = 95.21g   >>>answer