Question

If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (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

Answer: The actual yield of
`B_2O_3` is 60.0 g

Explanation:-

The balanced chemical reaction :

`B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)`

Mass of
`B_2H_6` =
`Density* Volume=1.131g/ml* 36.9ml=41.7g`

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} B_2H_6=(41.7g)/(27.668g/mol)=1.51moles`

According to stoichiometry:

1 mole of
`B_2H_6` gives = 1 mole of
`B_2O_3`

1.51 moles of
`B_2H_6` gives =
`(1)/(1)* 1.51=1.51` moles of
`B_2O_3`

Theoretical yield of
`B_2O_3=moles* {Molar mass}}=1.14mol* 69.62g/mol=79.3g`

Percent yield of
`B_2O_3`=
`75.7\%`

`\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}* 100`

`75.7\%=\frac{\text{Actual yield}}{79.3}* 100`

`{\text{Actual yield}}=60.0g`

Thus the actual yield of
`B_2O_3` is 60.0 g