As the mass number of the isotopes of hydrogen increases, the number of protons

Please help I need the answer now. Write a balanced equation for combustion of pentane in plenty supply of air and in limited su

Ratling [72]

Pentane burns in plenty of air: CO₂ and H₂O is produced.

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

Pentane burns in limited amount of air: CO or even C is produced along with H₂O.

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

<h3>Explanation</h3>

Pentane is a hydrocarbon. There are five carbon atoms in each of its molecule. Its molecular formula will be C₅H₁₂.

Hydrocarbon fuels burn to produce CO₂ when there's plenty of air.

? C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

Among all species in this reaction, C₅H₁₂ has the largest number of atoms per molecule. Assume that the coefficient of C₅H₁₂ is one.

1 C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

C₅H₁₂ is the only reactant that contains C atoms. There are 5 C atoms in a C₅H₁₂ molecule. There should be the same number of C atoms in the products.
CO₂ is the only product that contains C atoms. There are one C atom in each CO₂ molecule. 5 C atoms correspond to 5 CO₂ molecules.

1 C₅H₁₂ + ? O₂ → 5 CO₂ + ? H₂O

Similarly, C₅H₁₂ is the only reactant that contains H atoms. H₂O is the only product that contains H atoms. There are 12 H atoms in one C₅H₁₂ molecule, which corresponds to 6 H₂O molecules.

1 C₅H₁₂ + ? O₂ → 5 CO₂ + 6 H₂O

Both CO₂ and H₂O are products that contains O atoms. There are 5 × 2 + 6 × 1 = 16 O atoms in total in 5 CO₂ molecules and 6 H₂O molecules. The 16 O atoms on the product side corresponds to 8 O₂ molecules on the reactant side.

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

All coefficients shall be whole numbers. If there's any fraction in this equation, multiply both sides by the least common multiple of their denominators.

Hydrocarbon fuels burn to produce H₂O and CO when there's a limited supply of air. C (soot) might also be produced. Assuming that only CO is produced. Try to balance the equation using the same method.

1 C₅H₁₂ + 11/2 O₂ → 5 CO + 6 H₂O

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

Less O₂ is consumed for each mole of C₅H₁₂.

Consider: What would be the balanced equation when only C is produced?

<h3>Reference</h3>

"Products and effects of combustion", GCSE Chemistry (Single Science), BBC Bitesize.

5 0

3 years ago

What does the arrangement of the periodic table help scientists see?

lina2011 [118]

The relation between the elements electro negativity, atomic radius, and ionization energy levels.

7 0

4 years ago

When a certain amount of MgF2 is added to water, the freezing point lowers by 3.5° C. What was the molality of the

Leni [432]

The molality of the solution is obtained as 0.63 m.

<h3>What is the freezing point?</h3>

The freezing point is the temperature at which the liquid is converted into solid.

We know that;

ΔT = 3.5° C

K = 1.86° C/m

i = 3

m = ?

Thus;

ΔT = K m i

m = ΔT/K i

m = 3.5° C/ 1.86° C/m * 3

m = 0.63 m

Learn more about freezing point:brainly.com/question/3121416

#SPJ1

6 0

2 years ago

SHOW WORK

Makovka662 [10]

Answer:

1.51367e+10 inches

Explanation:

1 mile = 63360

63360 x 238900 = 15136704000

Hope this helped!

6 0

3 years ago

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

3 0

3 years ago