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

The percent yield of the reaction is 91.8 %

Explanation:

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

    .....(1)

Given mass of = 4.0 g

Molar mass of = 63.12 g/mol

Putting values in equation 1, we get:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

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

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of

So, 0.3125 moles of oxygen gas will react with = of

As, given amount of 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

So, 0.3125 moles of oxygen gas will produce = of water

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

Molar mass of = 69.93 g/mol

Moles of = 0.130 moles

Putting values in equation 1, we get:

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

Experimental yield of = 8.32 g

Theoretical yield of = 9.052 g

Putting values in above equation, we get:

Hence, the percent yield of the reaction is 91.8 %

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.