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SVETLANKA909090 [29]
3 years ago
15

Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . What is the theoretical yie

ld of carbon dioxide formed from the reaction of of methane and of oxygen gas?
Chemistry
2 answers:
avanturin [10]3 years ago
6 0

Completed question:

Gaseous methane  ( C H 4 )  reacts with gaseous oxygen gas  ( O 2 )  to produce gaseous carbon dioxide  ( C O 2 )  and gaseous water  ( H 2 O ) . What is the theoretical yield of carbon dioxide formed from the reaction of   0.16  g of methane and  0.83  g  of oxygen gas?

Answer:

0.44 g of CO₂

Explanation:

The reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

So, by the equation, we can notice that 1 mol of methane (CH₄) reacts with 2 moles of oxygen (O₂) to form 1 mole of carbon dioxide (CO₂) and 2 moles of water (H₂O).

The yield of a reaction is how much of the product is formed when a determined quantity of the reagents reacts. In the real-life, some complications can happen, such as parallel reactions, problems in pressure, temperature, or even in the reactor, and so the real yield will be less than the one predicted by the reaction (theoretical yield).

First, let's identify which one of the reagents is the limiting, which is, the one that is totally consumed, and so, will be the one that will conduct the yield.

The stoichiometry between the two reagents is 1 mol of CH₄ to 2 moles of O₂. The molar mass of CH₄ is 16 g/mol, and of O₂ is 32 g/mol, thus multiplying the number of moles by the molar mass, the ratio between them is:

16 g of CH₄/64 g of O₂

According to the Proust Law, this ratio must be constant. So, if there is 0.16 g of methane, the mass of oxygen must be:

16/32 = 0.16/m

16m = 0.16*32

16m = 5.12

m = 0.32 g of O₂

Because there is 0.83 g of O₂ it is in excess and CH₄ is the limiting reagent.

The stoichiometry between CH₄ and CO₂ is 1 mol of CH₄ to 1 mol of CO₂, the molar mass of CO₂ is 44 g/mol, so the mass rate is:

16 g of CH₄/44 g of CO₂

And the mass formed of CO₂:

16/44 = 0.16/x

16x = 44*0.16

16x = 7.04

x = 0.44 g of CO₂

maks197457 [2]3 years ago
4 0

Answer:

The theoretical yield of carbon dioxide formed from the reaction of methane and oxygen gas is 1 mol.

Explanation:

The equation below shows the reaction of combustion of methane:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

When no reactants are wasted and the reaction is complete, 1 mol of gaseous CO₂ will be produced. This means that the theoretical yield of carbon dioxide is 1 mol.

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3 years ago
A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a
svlad2 [7]

<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = 62.364\text{ L. Torr }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol

To calculate the mass from given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

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What information does DNA provide to all of the cell's organelles
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It gives each organelle in the cell its own specific function

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g Acetic acid is diluted with water to make a solution of vinegar. You have a sample of vinegar that contains 16.7 g of acetic a
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Answer:

0.278 mol

Explanation:

Step 1: Given and required data

Mass of acetic acid (m): 16.7 g

Chemical formula of acetic acid: CH₃COOH (C₂H₄O₂)

Step 2: Calculate the molar mass (M) of acetic acid

We will use the following expression.

M(C₂H₄O₂) = 2 × M(C) + 4 × M(H) + 2 × M(O)

M(C₂H₄O₂) = 2 × 12.01 g/mol + 4 × 1.01 g/mol + 2 × 16.00 g/mol = 60.06 g/mol

Step 3: Calculate the number of moles (n) of acetic acid

We will use the following expression.

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6 0
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If a balloon containing 3000 L of gas at 39°C and 99 kPa rises to an altitude where the pressure is 45.5 kPa and the temperature
ludmilkaskok [199]

\tt =3000~L\times \dfrac{289}{312}\times \dfrac{99}{45.5}

<h3>Further explanation</h3>

Given

3000 L of gas at 39°C and 99 kPa to 45.5 kPa and 16°C,

Required

the new volume

Solution

Combined with Boyle's law and Gay Lussac's law  

\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}

T₁ = 39 + 273 = 312

T₂ = 16 + 273 = 289

Input the value :

V₂ = (P₁V₁.T₂)/(P₂.T₁)

V₂ = (99 x 3000 x 289)/(45.5 x 312)

or we can write it as:

V₂ = 3000 L x (289/312) x (99/45.5)

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