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3 years ago

How do you write a balanced equation for the combustion ofbenzene?

Chemistry

1 answer:

forsale [732]3 years ago

4 0

Answer: The balanced chemical equation for the combustion of benzene is:

$2C_6H_6+15O_2\rightarrow 12CO_2+6H_2O$

Explanation:

Combustion reactions : It is defined as the reactions in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water as a product.

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

The balanced chemical equation for the combustion of benzene will be:

$2C_6H_6+15O_2\rightarrow 12CO_2+6H_2O$

By Stoichiometry of the reaction we can say that,

2 mole of benzene reacts with 15 moles of oxygen gas to produce 12 moles of carbon dioxide and 6 moles of water molecule.

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Trimix 10/50 is a gas mixture that contians 10% oxygen and 50% helium, and the rest is nitrogen. If a tank of trimix 10/50 has a

Nataly_w [17]

Answer: 1.61 x 10⁴ kPa

Dalton's law states that the sum of the partial pressures of each gas equals the total pressure of the gas mixture. According to this law,

Pi = xi P

where Pi is the partial pressure of the gas i, xi is the mole fraction of the gas i in the gas mixture and P is the total pressure.

The mole fraction is defined as the quotient between the moles of solute (ni) and the total moles of the mixture (nt), which is calculated by adding the moles of all its components:

xi = $\frac{n_{i} }{n_{t} }$

In the Trimix 10/50 mix you have 10% oxygen, 50% helium and 40% nitrogen.

To calculate the total number of moles of the mixture and thus determine the molar fraction of helium, we consider 100 g and calculate the number of moles that represent 10 g of O₂ (n₁), 50 g of He (n₂) and 40 g of N₂ (n₃):

n₁ = 10 g x $\frac{1 mol}{31.998 g}$ = 0.313 mol

n₂ = 50 g x $\frac{1 mol}{8.005 g}$ = 6.246 mol

n₃ = 40 g x $\frac{1 mol}{28.013 g}$ = 1.428 mol

Then the total number of moles (nt) will be:

nt = n₁ + n₂ + n₃ = 0.313 mol + 6.246 mol +1.428 mol

nt = 7,987 mol

Then, the mole fraction of helium (x₂) in the mixture will be,

x₂ = $\frac{6.246 mol}{7.987 mol}$ = 0.78

and the partial pressure of helium in the mixture, according to Dalton's law, will be:

P₂ = x₂ P = 0.78 x 2.07 x 10⁴ kPa

P₂= 1.61 x 10⁴ kPa

So, the partial pressure of helium if a tank of trimix 10/50 has a total pressure of 2.07 x 104 kPa is 1.61 x 10⁴ kPa

5 0

3 years ago

An illegal drug stimulates the parasympathetic nervous system, causing the user to appear tranquilized. What could indicate to a

Musya8 [376]

B. Low blood pressure and drooling. If the user is tranquilized, he would not be able to hinder the production of saliva, leading to drooling. And if the victim was tranquilized, his heart would slow down, not speed up, resulting in a lower blood pressure.

6 0

3 years ago

Which part of the food web is NOT a living thing?

stiks02 [169]

Answer:

the sun

Explanation:

the sun is not alive and plants use photosynthesis to eat the radiation emitted by the sun.

4 0

3 years ago

What is the oxidation state of cl in hclo4?

Irina-Kira [14]

There is 1 H atom: (1)(+1) = +1 The oxidation number of O is -2. There are 4 O atoms here: (4)(-2) = -8 So the oxidation state of Cl is +7.

NOTE: The maximum positive oxidation number for chlorine is +7, the same as its group number (VII).

3 0

3 years ago

Read 2 more answers

Lithium reacts with bromine (Br2) in a synthesis reaction to produce lithium bromide. Determine the limiting reactant if 25.0 gr

Lunna [17]

Answer: Bromine is the limiting reactant

Explanation:

First of all let's generate a balanced equation for the reaction

2Li + Br2 —> 2LiBr

Molar Mass of Li = 7g/mol

Molar Mass of Br2 = 2x80 = 160g/mol

From the question given, were told that 25g of Li and 25g Br2 were present at the take-off of the reaction. Converting these Masses to mole, we have:

Number of mole of Li = 25/7 = 3.6moles

Number of mole of Br2 = 25/160 = 0.156mol.

To know which is the limiting reactant, we have to compare the ratio of the number of mole of experimental Li and Br2 to that of theoretical Li and Br2

For the experimental yield:

Li : Br2 = 3.6/ 0.156 = 23 : 1

For the theoretical yield:

Li : Br = 2 : 1

From the above, we see clear that Br2 is the limiting reactant because according to the equation( which gives the theoretical yield), for every 2moles of Li, 1mole of Br2 is used up. But this is not so from the experiment conducted as 23moles required 1mole of Br2.

4 0

3 years ago