Explain the process of fading of a carpet upon repeated exposure to sunlight?

A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

Anit [1.1K]

Answer: The percent yield of the reaction is 32.34 %

Explanation:

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

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

For benzoic acid:

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

The chemical equation for the reaction of benzoic acid and methanol is:

$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

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

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

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

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

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

6 0

3 years ago

Dose your heart muscle get tired? Yes or no.

WARRIOR [948]

Answer:

no 

Explanation:

our heart muscles never get tired, because it has to pump blood in our body 72 times a minute, it is made of special cardiac muscles which helps it to perform it's function without getting tired ....

i hope it helped.....

5 0

2 years ago

What is the maximum amount of CO2

Wittaler [7]

Answer:

10.6 g CO₂

Explanation:

You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.

To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*

C₂H₂ + O₂ ----> 2 CO₂ + H₂

9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole

= 31.5 g CO₂

3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole

= 10.6 g CO₂

10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.

8 0

1 year ago

How many moles of Calcium Oxide are needed to produce 4 moles of Calcium Hydroxide?

Tpy6a [65]

Taking into account the reaction stoichiometry, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CaO + H₂O → Ca(OH)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CaO: 1 mole
H₂O: 1 mole
Ca(OH)₂: 1 mole

<h3>Moles of CaO required</h3>

The following rule of three can be applied: If by stoichiometric reaction 1 mole of Ca(OH)₂ is produced by 1 mole of CaO, 2 moles of Ca(OH)₂ are produced by how many moles of CaO?

$moles of CaO=\frac{2 moles of Ca(OH)_{2}x1 mol of CaO }{1 mole of Ca(OH)_{2}}$

moles of CaO= 2 moles

Finally, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

5 0

2 years ago

If 3.6 g of aluminum completely reacts, how much Al2O3 (in grams) can be produced

marissa [1.9K]

The amount in grams of Al₂O₃ produced is approximately 6.80 g.

Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:

AL + O₂ → Al₂O₃

Let's balance it

4AL + 3O₂ → 2Al₂O₃

4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,

Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,

Atomic mass of AL = 27 g

Molar mass of Al₂O₃ = 101.96 g/mol

4(27 g) of AL gives 2(101.96 g) of Al₂O₃

3.6 g of AL will give ?

cross multiply

mass of Al₂O₃ produced = 3.6 × 203.92 / 108 = 734.112 / 108 = 6.797

mass of Al₂O₃ produced = 6.80 g.

8 0

2 years ago