Hello can you please help me to solve above questions. .

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

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:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

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:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

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 = $\frac{2}{12}\times 0.3125=0.052mol$ 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 = $\frac{5}{12}\times 0.3125=0.130moles$ 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:

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

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

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

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:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

3 years ago

What would happen if less calcium chloride reacted and more unknown carbonate reacted?

Anit [1.1K]

A white insoluble solid would appeaer

8 0

3 years ago

Listed below are possible solutes. In your possession is a beaker with 500 milliliters of water. You want to make an aqueous sol

slavikrds [6]

Answer: I & III

Explanation: Solutes are the substances which are minimum in quantity and which is required to dissolve in the solvent (which is larger in quantity) in order to make a solution.

In the asked question, it is given that the water is the solvent and from the given solutes we have to pick which would make an aqueous solution with the highest concentration of solute possible.

Thus the most appropriate answers could be the Ammonia and hexanol which can make the highest possible concentration of solute as ammonia is the gas which is highly soluble in water and hexanol is an alcohol which has an affinity for water. Thus the correct option is I & III

8 0

3 years ago

Read 2 more answers

Atoms full valence electron shells are chemically stable. Most noble gases have eight valence electrons. So, one bottle for chem

kap26 [50]

Gain or lose.
The exchange of electrons in chemical bonding seeks to fulfill the octet rule. There are some exceptions, such as with hydrogen and helium, whose valence shells have a capacity of two electrons.

6 0

3 years ago

Read 2 more answers

How many moles of hydrogen are needed to produce 13.78 mol of ethane?

mamaluj [8]

Ethane is an alkane. Methane is also an alkane and is considered to be the simplest alkane. The difference is ethane has only 2 carbon. That carbon has 6 hydrogen attached to it. So what we do is we multiply the moles of ethane by the number of hydrogen (by dimension analysis) resulting to 82.68 moles H.

4 0

3 years ago

Read 2 more answers

Hello can you please help me to solve above questions. .​

Hello can you please help me to solve above questions. .