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lara31 [8.8K]
3 years ago
9

The combustion of methane is a reaction commonly used in chemistry problems due to its ability to fit into multiple topics. So i

t should not surprise you to see it here as well. How many L of CO2 would be produced if 45 g of CH4 was combusted with ample oxygen in a room that was 90 degrees Celsius and under 1 atm of pressure
Chemistry
1 answer:
lyudmila [28]3 years ago
8 0

Answer:

20.76 L OF CO2 WILL BE PRODUCED BY 45 G OF METHANE.

Explanation:

Equation of the reaction:

CH4 + 02 --------> CO2 + 2H20

Molar mass of methane = ( 12+ 1*4) g/mol = 16 g/mol

Calculate the number of moles present in 45 g of methane

1 mole of methane = 16 g / mol of methane

(45 / 16) mole of methane = 45 g of methane

= 2.8125 moles

Using the ideal gas equation:

PV = nRT

P = 1 atm

n = 2.812 moles

T = 90 C

R = 0.082 L atm/ mol C

V = unknown

So we have:

V = nRT / P

V = 2.8125 * 0.082 * 90 / 1

V = 20.756 L

In the production of CO2 by 45 g of methane, 20.756 L of methane was used.

Then, the volume of CO2 produced by this volume will be 20.756 L since 1 mole of methane produces 1 mole of CO2.

In other words;

1 mole of CH4 = 1 mole of CO2

22.4 dm3 of CH4 = 22.4 dm3 of CO2

20.76 DM3 = 20.76 dm3

The volume of CO2 produced will therefore be 20.76 L

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Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi
Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g)     ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

[]=\frac{P}{R.T}

<em>Calculate ΔG at 25°C given the following sets of partial pressures.</em>

<em>Part A  130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.</em>

[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M

[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

<em>Part B  5.0atm SO₂, 3.0atm O₂, 30atm SO₃  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M

[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M

[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

<em>Part C Each reactant and product at a partial pressure of 1.0 atm.  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0
2 years ago
A beaker in your laboratory drawer has an inside diameter of 6.8 cm and a height of 8.9 cm. Using the equation V= arh, calculate
olga_2 [115]

Answer:

323.22 ml

Explanation:

Given that :

Diameter, d = 6.8cm

Height, h = 8.9cm

V = arh

Recall :

Volume, V = πr²h

Radius, r = diameter / 2 = 6.8 / 2 = 3.4cm

V = π * 3.4^2 * 8.9

V = 323.21961 cm³

Recall:

1ml = 1cm³

Hence,

323.21961 cm³ = 323.21961 ml

Volume = 323.22 ml

4 0
2 years ago
What is the reaction type? ​
Wewaii [24]

Answer:

exothermic reaction

Explanation:

3 0
2 years ago
Determine the molarity of 4.67 moles of Li2SO3 dissolved to make 2.04 liters of solution
PSYCHO15rus [73]
You take the moles divided by the liters to get the molarity.
8 0
3 years ago
Three isotopes of oxygen are oxygen-16 oxygen-17 and oxygen-18. Write the symbol for each, including the atomic number and mass
notsponge [240]
<span>The notation of the isotopes using the atomic number and the mass number consists of the symbol of the atom, preceded by the mass number as a superscript and the atomic number as a superscript.

All the isotopes of the same element have the same atomic number. They only vary the mass number.

So, all the isotopes of oxygen have atomic number 8.

The isotope oxygen-16 has mass number 16, so it is written with the symbol O preceded by the number 16 as a superscript and the number 8 as a subscript (the two numbers to the right of the chemical symbol).

The isotope oxygen-17 has mass number 17, so it is written with the symbol O preceded by the number 17 as a superscript and the number 8 as a subscript.

The isotope oxygen-18 has mass number 18, so it is written with the symbol O preceded by the number 18 as a superscript and the number 8 as a subscript.</span>
4 0
3 years ago
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