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Firlakuza [10]
4 years ago
6

Suppose 0.410 kg of hexane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0 °C. Calculate the volume o

f carbon dioxide gas that is produced.Round your answer to 3 significant digits.
Chemistry
1 answer:
3241004551 [841]4 years ago
8 0

Answer:

The answer is 671 litres of carbon dioxide is produced from 0.410 kg of hexane

Explanation:

We first write a balanced reaction for the complete combustion of hexane thus

The stoichiometry of the cumbustion of hexane in air is

2C6H14(g)+18O2(g)→12CO2(g)+14H2O(l) or

C6H14(g)+9O2(g)→6CO2(g)+7H2O(l)

From the above reaction it is observed that one mole of hexane burns completely in the presence of oxygen to produce 6 moles of carbon dioxide

Therefore we calculate the nuber of moles of hexane present in the sample thus

Mass hexane of sample = 0.41 kg

Molar nass of hexane = 86.18 g/mol

number of moles of hexane = (mass of hexane)/(molar mass of hexane) = (0.41×1000)/86.16 = 410/86.16 = 4.76 moles

As we have seen from the chemical reaction, 1 mole of H6H14 produces 6 moles of CO2 hence 4.76 moles of Hexane produces

4.76×6 moles of CO2 which is 28.55 moles of CO2

From the question we have the temperature and the pressure of the production of CO2 as

Temperature of reaction = 13° C converting to kelving gives= 13+273.15 = 286.15 K

and pressure = 1 atmosphere or 101325 Pa

13.0∘C=13.0∘C+273.15=286.15 K

The volume of the produced CO2 can be calculated using the combined ideal gas equation given by

P×V=n×R×T where

Here

P = Gas pressure (of CO2 )

V = Volume (of the CO2)

n = number of moles of gas (CO2) present

R = universal gas constant, equal to 0.0821 atm× L/(mol× K )

T = absolute temperature in Kelvin

Thus we have

1×V = 28.55×0.0821×286.15  or V  = 670.76L

Rounding up the answer to 3 significant digits we have

670.76L ≅ 671L

671 litres of carbon dioxide is produced from 0.410 kg of hexane

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Which type of reaction is the Haber process: N2(g) + 3 H2(g) → 2 NH3(g) + heat? *
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exothermic, with a decrease in entropy

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3 years ago
A mixture of 0.2000 mol of CO2, 0.1000 mol of H2 and 0.1600 mol of H2O is placed in a 2.000-L vessel. The following equilibrium
Slav-nsk [51]

Answer:

a) p(CO2) = 4.103 atm

p(H2) = 2.0515 atm

p(H2O) = 3.2824 atm

b) pCO2  = 3.8754 atm

pH2 =  1.8239 atm

pCO = 0.2276 atm

pH2O = 3.51 atm

c) Kp= 0.113

Explanation:

Step 1: Data given

Moles of CO2 = 0.2000 mol

Moles of H2 = 0.1000 mol

Moles of H2O = 0.1600 mol

Volume = 2.000 L

Temperature = 500 k

Step 2: The balanced equation

CO2 (g) + H2 (g) → CO (g) + H2O (g)

Step 3: Calculate the initial partial pressures of CO2, H2, and H2O.

pV = nRT

P = (nRT)/V

p(CO2) = (0.2000 mol * 0.08206 * 500)/2.000L

⇒ p(CO2) = 4.103 atm

p(H2) = (0.1000 mol * 0.08206 * 500)/2.000L

⇒ p(H2) = 2.0515 atm

p(H2O) = (0.1600 mol * 0.08206 * 500)/2.000L

⇒ p(H2O) = 3.2824 atm

b. At equilibrium PH2O= 3.51 atm. Calculate the equilibrium partial pressures of CO2, H2, and CO.

Step 1: Calculate the change in pH2O

The change in pH2O = 3.51 - 3.2824 = 0.2276

Step 2: the initial pressures

pCO2 = 4.103 atm

pH2 = 2.0515 atm

pCO = 0 atm

pH2O = 3.2824

Step 3: The partial pressure at the equilibrium

Since there reacts 0.2276 atm for H2O, and the mol ratio is 1:1:1:1

For each gas, there will react 0.2276 atm

pCO2 = 4.103 - 0.2276 = 3.8754 atm

pH2 = 2.0515 - 0.2276 = 1.8239 atm

pCO = 0 + 0.2276 = 0.2276 atm

pH2O = 3.51 atm

c. Calculate Kp for this reaction

Kp = (pCO * pH2O)/ (pCO2 * pH2)

Kp = (0.2276 * 3.51) /( 3.8754 *1.8239)

Kp = 0.113  

5 0
3 years ago
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