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4 years ago

6. 100 ml of gaseous hydrocarbon consumes 300

Chemistry

1 answer:

mario62 [17]4 years ago

7 0

Answer:

a. C₂H₄

Explanation:

At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).

Hence, the complete combustion reaction that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.

Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).

A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.

a. C₂H₄:

C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)

Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.

The following analysis just shows that the other options are not right.

b. C₂H₂:

2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)

The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.

с. С₃Н₈

C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)

The mole ratio is 1 mol C₃H₈ : 5 mol O₂

d. C₂H₆

2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)

The mole ratio is 2 mol C₂H₆ : 7 mol O₂

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What mass of zinc (molar mass 65.4 g moll) does it take to produce 0.50 mole of H2(g)?

Hatshy [7]

Answer:

32.7 g of Zn

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Zn + 2HCl —> ZnCl₂ + H₂

From the balanced equation above,

1 mole of Zn reacted to produce 1 mole of H₂

Next, we shall determine the number of mole of Zn required to produce 0.5 mole of H₂. This can be obtained as follow:

From the balanced equation above,

1 mole of Zn reacted to produce 1 mole of H₂.

Therefore, 0.5 mole of Zn will also react to produce to 0.5 mole of H₂.

Thus, 0.5 mole of Zn is required.

Finally, we shall determine the mass of 0.5 mole of Zn. This can be obtained as follow:

Mole of Zn = 0.5 mole

Molar mass of Zn = 65.4 g/mol

Mass of Zn =?

Mass = mole × molar mass

Mass of Zn = 0.5 × 65.4

Mass of Zn = 32.7 g

Thus, 32.7 g of Zn is required to produce 0.5 mole of H₂.

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3 years ago

At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

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3 years ago

Find the mass of 1.112 moles HF?

Alexeev081 [22]

Answer:

22.25 g

Explanation:

To find the mass, you need to convert moles to grams and get 22.25 g.

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3 years ago

Ethanol (CH3CH2OH), the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuri

Fed [463]

Answer:

a) 88.48%

b) 0.05625 mol

Explanation:

2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1

CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2

a) CH₃CH₂OH = 46.0684 g/mol

CH₃CH₂OCH₂CH₃ = 74.12 g/mol

1 mol CH₃CH₂OH ______ 46.0684 g

x ______ 50.0 g

x = 1.085 mol CH₃CH₂OH

1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g

y ______ 35.9 g

y = 0.48 mol CH₃CH₂OCH₂CH₃

100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃

w _____ 0.48 mol CH₃CH₂OCH₂CH₃

w = 88.48%

b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.

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3 years ago