What is mass in g of 22 moles of copper?

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

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

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

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

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

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

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

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

3 years ago

The name 3-chloro-2-propylbutane is not an accepted name for the compound described. what would be the correct name for this com

Illusion [34]

The longest hydrocarbon chain in the given compound is hexane, therefore it is the parent chain to be considered with one methyl group attached to the 3rd carbon and one chloro attached on the 2nd carbon, therefore the name of the compound is 2-chloro-3-methylhexane

3 0

3 years ago

Choose the false statement regarding resistance

Dmitry_Shevchenko [17]

I would say all if no statements are available

4 0

3 years ago

Solid lead IV oxide decomposes into solid lead and oxygen gas.

vredina [299]

The balanced equation for the decomposition of solid lead iv oxide is as follows: 2PbO2 = 2PbO + O2.
Lead IV oxide decompose to give lead ll oxide and oxygen. Lead iv oxide is thermally unstable and it usually decomposes into oxygen and lead ll oxide when heated. Lead ll oxide is more stable than lead lV oxide.

7 0

3 years ago

2 bromocyclopentanamine structural formula

Elis [28]

Answer:

The structure is given in attached file.

Explanation:

Explanation

2-bromocyclopentamine (Figure attached) is a synthetic compound which is synthesized by substitution reaction of cyclopentamine and hydrobromide. Its molecular formula and molecular mass are C5H10NBr and 164.05 mol/g respectively. It is a very reactive compound so it doesn’t available in pure form, it is present in market as a mixture of 2-bromocyclopentamine and Hydrobromide.

Properties :

Its boiling point is 115 0C

Its melting point is – 75 oC

It is highly flammable

It is highly toxic

It is irritant

It is corrosive in nature

5 0

3 years ago