Which of the following is one of the ending compounds

What volume of air is needed to burn an entire 55-L (approximately 15-gal) tank of gasoline? Assume that the gasoline is pure oc

White raven [17]

Answer:

The volume of air required is 527,686.25L.

Explanation:

When the question says "burn", it refers to a combustion reaction, where a substance (in this case octane) reacts with oxygen to produce carbon dioxide and water.

Step 1: Write a balanced equation

Considering it is a combustion reaction, the balanced equation is:

C₈H₁₈ + 12.5 O₂ ⇄ 8 CO₂ + 9 H₂O

In this step, we start balancing elements that are present only in one compound on each side of the equation, namely, carbon and hydrogen.

To finish, it is important to count that there are the same number of atoms on both sides of the equation. In this case there are 8 atoms of Carbon, 18 atoms of Hydrogen and 25 atoms of Oxygen, so it is balanced.

Step 2: Find out the mass of C₈H₁₈

Since the balanced equation gives us information about the mass of C₈H₁₈ involved in the reaction, we need to find out how many grams we have.

The info we have is:

55 L of gasoline (assuming gasoline to be pure octane).
The density of octane is 0,70 g/cm³

Density relates mass and volume, so we can find out how many grams are represented by 55 L. Since the units used are different, first we need to convert liters into cm³. We use the conversión factor 1 L = 1000 cm³.

$55L.\frac{1000cm^{3} }{1L} =55000cm^{3}$

Since density = mass/volume, we can solve for mass:

$mass = density.volume=0.70\frac{g}{cm^{3} } .55,000cm^{3} =38,500g$

Step 3: Establish the theoretical relationship between the mass of octane and the moles of oxygen.

This relationship comes from the balanced equation.

For octane:

molar mass of C₈H₁₈ = molar mass of C . 8 + molar mass of H . 18 =

12 g/mol . 8 + 1g/mol . 18 = 114 g/mol

According to the balanced equation reacts 1 mol of octane, which means 114 grams of it.

For oxygen:

According to the balanced equation, 12.5 moles of oxygen react.

Then, the relationship is 114 g octane : 12.5 moles of oxygen

Step 4: Use the theorethical relationship to find the moles of oxygen that reacted

We use the mass of octane found in step 2 and apply the proper conversión factor.

$38,500g (octane) . \frac{12.5mol (oxygen)}{114g (octane)} = 4,221.49mol(oxygen)$

Step 5: Find out the volume of oxygen.

We know that 1 mol of any gas at room temperature occupies about 25 L. Then,

$4,221.49mol(oxygen).\frac{25L(oxygen)}{1mol(oxygen)} =105,537.25 L (oxygen)$

Step 6: Look the volume of air that contains such amount of oxygen

Given oxygen represents 20% of air, we can use that relationship to find the volume of air.

$105,537.25L(oxygen).\frac{100L(air)}{20L(oxygen)} =527,686.25L (air)$

4 0

4 years ago

3. Description of which organelle betul "Creates a physical barrier ;

Ghella [55]

The cell membrane performs those functions. It is the outer layer of a cell and it creates a barrier, regulates material entry and exit, uses chemical signals, and creates structure.

4 0

3 years ago

Which of the following steps will slow down the reaction between coal and oxygen?

Nezavi [6.7K]

Decrease the temperature of the oxygen because if coal goes air born it can explode and wipe out power plants like they were not there

7 0

3 years ago

Read 2 more answers

Which force pulls an airplane toward the Earth and is related to the mass of the airplane?

Zolol [24]

EASY gravity is the answer lol

4 0

3 years ago

Read 2 more answers

For the following example, identify the following:

guapka [62]

Answer: Option (A) is the correct answer.

Explanation:

According to the Gibb's free energy conditions:

If $\Delta G$ < 0, then reaction is spontaneous.

If $\Delta G$ > 0, then reaction is non-spontaneous.

If $\Delta G$ = 0, then reaction is equilibrium.

Since, the given reaction is $F_{2}(l) \rightarrow F_{2}(g)$ . It denotes that liquid state of fluorine is changing into gaseous state. And, the boiling point of fluorine is very low.

Therefore, at low and high temperature fluorine is a gas. Hence, the reaction is spontaneous in nature.

Thus, we can conclude that the statement at low temperature, the reaction is spontaneous and $\Delta G$ < 0 and at high temperature, the reaction is spontaneous and $\Delta G$ < 0, is correct.

4 0

3 years ago