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3 years ago

What is the volume of 33.25g of butane gas at 293 C and 10.934 kPa?

Chemistry

1 answer:

PSYCHO15rus [73]3 years ago

6 0

Answer:

V = 240.79 L

Explanation:

Given data:

Volume of butane = ?

Temperature = 293°C

Pressure = 10.934 Kpa

Mass of butane = 33.25 g

Solution:

Number of moles of butane:

Number of moles = mass/ molar mass

Number of moles = 33.25 g/ 58.12 g/mol

Number of mole s= 0.57 mol

Now we will convert the temperature and pressure units.

293 +273 = 566 K

Pressure = 10.934/101 = 0.11 atm

Volume of butane:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

V = nRT/P

V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm

V = 26.49 L/0.11

V = 240.79 L

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The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at

horsena [70]

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Bringing out the parameters mentioned in the question;

Vapor pressure = 94.4 mm of Hg

The vaporization reaction is given as;

C₆H₆(l) ⇄ C₆H₆(g)

Equilibrium in terms of activities is given by:

K = a(C₆H₆(g)) / a(C₆H₆(l))

Activity of pure substances is one:

a(C₆H₆(l)) = 1

Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg standard pressure

We now have;

K = 94mmHg / 760mmHg = 0.12421

Gibbs free energy is given as;

ΔG = - R·T·ln(K)

where R = gas constant = 8.314472J/molK

So ΔG° of vaporization of benzene is:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol

Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

Hence:

ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

6 0

4 years ago

In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n

artcher [175]

Stoichiomety:

1 moles of C + 1 mol of O2 = 1 mol of CO2

multiply each # of moles times the atomic molar mass of the compund to find the relation is weights

Atomic or molar weights:

C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol

Stoichiometry:

12 g of C react with 32 g of O2 to produce 44 g of CO2

Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen

And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.

You cannot obtain 72 g of CO2 from 18 g of C.

May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.

3 0

4 years ago

Read 2 more answers

4.8 tons of pig ovaries were required to extract 12.0 mg of estrogen

bearhunter [10]

Is this a question? if so please elaborate.

7 0

3 years ago

Calculate the pH of 1.00 X 10^-6 M HCI *

murzikaleks [220]

Answer: 6

Explanation:

To find pH you have to do -log(concentration of H+)

7 0

3 years ago

How many moles of oxygen atoms are present in 30.5 grams of hydrogen peroxide (h2o2)?

Pepsi [2]

1.8 moles of oxygen atoms are present in 30.5 grams of hydrogen peroxide.

<u>Explanation:</u>

First we have to convert the given weight of hydrogen peroxide to molar mass of hydrogen peroxide. So for this, we have to divide the given weight with the molecular mass of hydrogen peroxide.

$\text {Molecular mass of } \mathrm{H}_{2} \mathrm{O}_{2}=(2 \times 1)+(2 \times 16)=2+32=34 \mathrm{g}$

So,

$\text {Molar mass of } \mathrm{H}_{2} \mathrm{O}_{2}=\frac{30.5}{34}=0.90 \text { moles of } \mathrm{H}_{2} \mathrm{O}_{2}$

Second step, in this moles, 2 molecules of oxygen are present. Thus 1 mole of Hydrogen peroxide consists of 2 moles of oxygen. Then,

$0.90 moles of $\mathrm{H}_{2} \mathrm{O}_{2}=2 \times 0.90=1.8$ moles of oxygen$

So, 30.5 grams of hydrogen peroxide consists of 1.8 moles of oxygen.

4 0

3 years ago