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3 years ago

What is the percent composition of hydrogen (H) in benzene (C6H6)?

Chemistry

2 answers:

Scorpion4ik [409]3 years ago

8 0

Answer : The percent composition of hydrogen in benzene is, 7.69 %

Explanation: Given,

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

First we have to calculate the molar mass of benzene.

Molar mass of benzene $(C_6H_6)$ = $6(12)+6(1)=78g/mole$

Now we have to calculate the percent composition of hydrogen in benzene.

As we now that there are 6 number of carbon atoms and 6 number of hydrogen atoms.

The mass of hydrogen = $6\times 1=6g$

Formula used :

$\%\text{ Composition of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Molar mass of hydrogen}}\times 100$

Now put all the given values in this formula, we get the percent composition of hydrogen in benzene.

$\%\text{ Composition of hydrogen}=\frac{6}{78}\times 100=7.69\%$

Therefore, the percent composition of hydrogen in benzene is, 7.69 %

PtichkaEL [24]3 years ago

6 0

That's the answer on that picture

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Two molecules of one reactant combine with 3 molecules of another to produce 5 molecules of a product.

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Answer:

A = 2A + 3B → 5C

Explanation:

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2A + 3B → 5C

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3 years ago

According to Charles's law, what will happen to the volume of these balloons as temperature

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The volume will decrease and the balloons will be smaller

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2 years ago

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r

spin [16.1K]

Answer: The theoretical yield of acetanilide is 6.5 grams.

Explanation:

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

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

For aniline:

Given mass of aniline = $4.50\times 10^0=4.50g$ (We know that: $10^0=1$ )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

$\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol$

For acetic anhydride:

To calculate the mass of acetic anhydride, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of acetic anhydride = $(1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL$

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

$1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g$

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol$

The chemical equation for the reaction of aniline and acetic anhydride follows:

$C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH$

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = $\frac{1}{1}\times 0.048=0.048mol$ of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = $\frac{1}{1}\times 0.048=0.048mol$ of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

$0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g$

Hence, the theoretical yield of acetanilide is 6.5 grams.

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BartSMP [9]

Answer:

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Each liter of air has a mass of 1.80 grams. How many liters of air are contained in

natita [175]

Question:

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