All categories

3 years ago

If 1.3 L of C3H8 combusts according to the equation below, how much CO2 will be produced?

Chemistry

1 answer:

SCORPION-xisa [38]3 years ago

6 0

Answer:

0.162 moles of CO₂ are produced by this reaction

Explanation:

The reaction is:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) +4H₂O(g)

As we have the volume of propane, we need to know the mass that has reacted, so we apply density's concept.

Density = Mass / Volume → Density . Volume = Mass

0.00183 g/mL . 1300 mL = Mass → 2.379 g

We determine the moles → 2.379 g . 1mol / 44 g = 0.054 moles

Ratio is 1:3. 1 mol of propane can produce 3 moles of dioxide

Then, 0.054 moles may produce (0.054 .3)/1 = 0.162 moles

You might be interested in

Most people like to go outside on a sunny day, but too much sun exposure can cause sunburns as well as long term health problems

ss7ja [257]

Earths atmosphere protects you from the sun

4 0

3 years ago

Read 2 more answers

I am a gas. I am in the noble gas family and row 1. I have only 2 valence electrons I glow red-orange when in an electric field.

belka [17]

The element is Helium

Goodluck!

7 0

2 years ago

Read 2 more answers

What molecule is the ultimate source of hydrogen ions that are secreted into the gastric juice?

sladkih [1.3K]

Answer: carbonic acid

Explanation:

8 0

3 years ago

5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time

11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}$

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142$

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen = $d_{O_2}=\frac{1 m}{t}$

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane= $d_{CH_4}=\frac{y }{t}$

$\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142$

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0

3 years ago

Landfills are one source of CO2. As garbage decomposes, it releases CO2 into the air. Can you think of at least

Margaret [11]

Answer:

Recycling and reuse of materials

Explanation:

One of the greatest problems facing the human population is the problem of solid waste disposal. The menace of solid waste disposal has led to the idea of landfills. Land fills are depressions on the earth surface prepared for the purpose of solid waste disposal.

The most important approach towards solid waste disposal is the idea of recycling of materials. A material can be collected after use and processed into the same material or serve as a precursor in another manufacturing process. This means that no waste is generated as the materials which are supposed to be disposed of as solid waste are processed into other useful materials. This will reduce the volume of solid wastes generated that may need to be disposed in a landfill.

7 0

2 years ago