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 %

if you're asking what comes out, it's steam.