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ohaa [14]
1 year ago
9

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning

completely 1.00 L of octane (C8H18, density = 0.7025 g/mL), yielding carbon dioxide and water?
Chemistry
1 answer:
Temka [501]1 year ago
0 0

Answer:

V = 8963 L

Explanation:

Our strategy here is to realize this problem involves stoichiometric calculation based on the balanced chemical equation for the combustion of octane:

C₈H₁₈ + 25/2 O₂ ⇒ 8 CO₂ + 9 H₂O

From the density of octane we can obtain the number of moles:

D = m/V ⇒ m= D x V = 0.7025 g/mL x ( 1000 mL) = 702.5 g

MW octane  = 702.5 g/ 114.23 g/mol = 6.15 mol

Required  mol oxygen to react with octane:

6.15 mol octane x  25/2 mol O₂ / mol octane = 76.8 mol O₂

Now mol fraction is given by mol O₂ / total number mol air ⇒

n air = 76.8 mol O₂ / (0.2095 mol O₂ / mol air )  = 366.59 mol air

and from the ideal gas law we can compute the volume of air:

PV = nRT  ⇒ V = nRT/P

V = 366.59 mol air x 0.08205 Latm/Kmol x (25+ 273) K/ 1 atm

   = 8,963 Lts

Note we treat here  air as a compund which is allowed in combustion problems.

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A sample of gas initially occupies 3.35 L at a pressure of 0.950 atm at 13.0oC. What will the volume (in L) be if the temperatur
Marat540 [252]

Answer: V= 3.13 L

Explanation: solution attached:

Use combine gas law equation:

P1 V1 / T1 = P2 V2/ T2

Derive to find V2

V2 = P1 V1 T2 / T1 P2

Convert temperatures in K

T1= 13.0°C + 273 = 286 K

T2= 22.5°C + 273 = 295.5 K

Substitute the values.

4 0
2 years ago
Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the
Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by \eta.

The given data is as follows.

     \eta = 0.82,       T_{1} = (21 + 273) K = 294 K

     P_{1} = 200 kPa,     P_{2} = 1000 kPa

Therefore, calculate the final temperature as follows.

         \eta = \frac{T_{2} - T_{1}}{T_{2}}    

         0.82 = \frac{T_{2} - 294 K}{T_{2}}    

          T_{2} = 1633 K

Final temperature in degree celsius = (1633 - 273)^{o}C

                                                            = 1360^{o}C

Now, we will calculate the entropy as follows.

       \Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}

For 1 mole,  \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

It is known that for NH_{3} the value of C_{v} = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

     \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

                = 0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}

                = 0.0346 kJ/mol

or,             = 34.6 J/mol             (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0
2 years ago
Please answer these. It's part of Quantitave Chemistry. Calculating limiting raegents
Arada [10]

The limiting reactant refers to the reactant that is in small amount. The extent of the reaction depends on the amount of the limiting reactant.

<h3>What is limiting reactant?</h3>

The term limiting reactant refers to the reactant that is in small amount, We shall now solve the questions individually.

1) The reaction equation is;

CaO + H2O → Ca(OH)2

Number of moles of CaO = 10 g/56 g/mol = 0.18 moles

Number of moles of H2O = 10 g/18 g/mol = 0.56

Since the reaction is 1:1, CaO is the limiting reactant

Mass of calcium hydroxide = 0.18 moles * 74 g/mol = 13.32 g

2) The reaction equation is;

Mg + Br2 → MgBr2

Number of moles of Mg = 1 g/24 g/mol = 0.042 moles

Number of moles of Br2= 5 g/160 g/mol = 0.031 moles

Since the reaction is 1:1, Br2 is the limiting reactant

Mass of MgBr2 =  0.031 moles * 184 g/mol = 5.7 g

3) The reaction equation is;

CuO + H2 → Cu + H2O

Number of moles of CuO= 2.00 g/80 g/mol = 0.025 moles

Number of moles of H2 = 1.00g/2 g/mol = 0.5 moles

Since the reaction is 1:1, CuO is the limiting reactant

Mass of Cu = 0.025 moles * 63.5 g/mol = 1.59 g

4) The reaction equation is;

2Na + F2 → 2NaF

Number of moles of Na = 2.30 g/23 g/mol = 0.10 moles

Number of moles of F2 =  2.85 g/38 g/mol = 0.075 moles

If 2 moles of Na react with 1 mole of F2

x moles of  Na reacts with 0.075 moles of F2

x = 0.15

Hence Na is the limiting reactant

Mass of NaF = 0.10 moles * 42 g/mol = 4.2 g

5) The equation of the reaction is;

Fe2O3 + 2Al → 2Fe + Al2O3

Number of moles of Fe2O3  =  8.00 g/160 g/mol = 0.05 moles

Number of moles of Al = 2.16 g/27 g/mol = 0.08 moles

If 1 mole of Fe2O3  reacts with 2 mole of Al

x moles of Fe2O3 reacts with 0.08 moles of Al

x = 0.04 moles

Hence, Fe2O3 is the limiting reactant

1 mole of Fe2O3  yileds 2 moles of Fe

0.05 moles yileds 0.05 moles  * 2 moles/1 mole = 0.1 moles

Mass of Fe = 0.1 moles * 56 g/mol = 5.6 g

6) The equation of the reaction is;

2Al + 3Cl2 → 2AlCl3

Number of moles of Al = 13.5 g/27 g/mol = 0.5 moles

Num ber of moles of Cl2 =  42.6 g /71 g/mol = 0.6 moles

If 2 moles of Al reacts with 3 moles of Cl2

x moles of Al reacts with 0.6 moles of Cl2

x = 0.4 moles

Al is the limiting reactant

Mass of AlCl3 = 0.4 moles * 133 g/mol = 53.2 g

Leran more about limiting reactants: brainly.com/question/14225536

4 0
10 months ago
Charcoal (burned wood) that was used to make prehistoric drawings on cave walls in france was scraped off and analyzed. the resu
Finger [1]
TLDR: the answer is C. 22,920 years.

Half-life describes the amount of time for a radioactive substance to decay to one-half of the original substance’s weight. So, if we had 100g of C-14, after 5,730 years, only 50g remain; after another 5,730 years, only 25g would remain, and so on.

In this problem, we are meant to assume that the original amount of C-14 was 64g, and that, through decay, it forms N-14. We can figure out how many half lives have passed by figuring out how much 4 is out of 64 by dividing 64 by two repeatedly. Each time, count a half life.

64 - 32 (1 HL) - 16 (2 HL) - 8 (3 HL) - 4 (4 HL)

In this problem, 4 half lives have passed. We can now multiply this by the time for one half life to find how many years have passed.

4 x 5,730 = 22,920 years

Approximately 22,920 years have passed since the drawing was created.
6 0
1 year ago
What are the missing coefficients for the skeleton equation below?
Vika [28.1K]

Answer:

C. Double Replacement

Explanation:

A. Wrong because SR uses a compound and element. This equation is a compound and compound.

B. Wrong because it does not have O2 in the formula. All combustion reactions must have O2.

C. Correct because it is a compound reacting with a compound.

D. Wrong because the reactants did not form a single product. ex. (x + y > xy)

E. Wrong because the equation did not start with a single compound and break down. ex. (xy > x + y)

3 0
2 years ago
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