Question

Find percent yield:

The mass of B2O3 produced by the reaction of 4.00 g of B5H9, and 10.00 g of
O2 is 8.32 g. What is the percent yield?
2 B3H9 + 12 O2 => 5 B2O3 +9 H2O

87.2
92.8
91.8
75.5
74.5

Answer 1

Answer: The percent yield of the reaction is 91.8 %

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For
  `B_5H_9` :

Given mass of
`B_5H_9` = 4.0 g

Molar mass of
`B_5H_9` = 63.12 g/mol

Putting values in equation 1, we get:

`\text{Moles of }B_5H_9=(4g)/(63.12g/mol)=0.0634mol`

• For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of oxygen gas}=(10g)/(32g/mol)=0.3125mol`

The chemical equation for the reaction of
`B_5H_9` and oxygen gas follows:

`2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O`

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of
`B_2H_5`

So, 0.3125 moles of oxygen gas will react with =
`(2)/(12)* 0.3125=0.052mol` of
`B_2H_5`

As, given amount of
`B_2H_5` is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of
`B_2O_3`

So, 0.3125 moles of oxygen gas will produce =
`(5)/(12)* 0.3125=0.130moles` of water

Now, calculating the mass of
`B_2O_3` from equation 1, we get:

Molar mass of
`B_2O_3` = 69.93 g/mol

Moles of
`B_2O_3` = 0.130 moles

Putting values in equation 1, we get:

`0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol* 69.63g/mol)=9.052g`

To calculate the percentage yield of
`B_2O_3`, we use the equation:

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

Experimental yield of
`B_2O_3` = 8.32 g

Theoretical yield of
`B_2O_3` = 9.052 g

Putting values in above equation, we get:

`\%\text{ yield of }B_2O_3=(8.32g)/(9.052g)* 100\\\\\% \text{yield of }B_2O_3=91.8\%`

Hence, the percent yield of the reaction is 91.8 %