Question

4. Use the following reaction: C4H9OH + NaBr + H2SO4 C4H9Br + NaHSO4 + H2O If 15.0 g of C4H9OH react with 22.4 g of NaBr and 32.7 g of H2SO4 to yield 17.1 g of C4H9Br, what is the percent yield of this reaction?

Answer 1

Answer:

here u go

Explanation:

Use the following reaction: C4H9OH + NaBr + H2SO4 C4H9Br + NaHSO4 + H2O If 15.0 g of C4H9OH react with 22.4 g of NaBr and 32.7 g of H2SO4 to yield 17.1 g of C4H9Br, what is the percent yield of this reaction? Remember: percent yield is your (experimental yield/theoretical yield)x100.

Answer: The percent yield of the reaction is 61.5 %.

Explanation:

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

     .....(1)

For butanol:

Given mass of butanol = 15.0 g

Molar mass of butanol = 74 g/mol

Putting values in equation 1, we get:

For NaBr:

Given mass of NaBr = 22.4 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

For sulfuric acid:

Given mass of sulfuric acid = 32.7 g

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

For the given chemical reaction:

As, the reactants are present in 1 : 1 : 1 ratio. So, the reactant having minium number of moles will be considered as the limiting reagent.

Here, the limiting reagent is butanol because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of butanol produces 1 mole of bromobutane

So, 0.203 moles of butanol will produce =  of bromobutane

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

Molar mass of bromobutane = 137 g/mol

Moles of bromobutane = 0.203 moles

Putting values in equation 1, we get:

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

Experimental yield of bromobutane = 17.1 g

Theoretical yield of bromobutane = 27.81 g

Putting values in above equation, we get:

Hence, the percent yield of the reaction is 61.5 %.