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2 years ago

Octane has a density of 0.703 g/ml. Calculate the mass of CO2(g) produced by burning one

Chemistry

1 answer:

STatiana [176]2 years ago

8 0

The mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

The given parameters:

Density of the octane, ρ = 0.703 g/ml
Volume of octane, v = 3.79 liters

The mass of the octane burnt is calculated as follows;

$m = \rho V\\\\m = 0.703 \ \frac{g}{ml} \times 3.79 \ L \ \frac{1000 \ ml}{L} \\\\m = 2,664.37 \ g$

The combustion reaction of octane is given as;

$2C_8H_{18} + \ 25O_2 \ --> \ 16CO_2 \ + \ 18H_2O$

From the reaction above:

228.46 g of octane -------------------> 704 g of CO₂ gas

2,664.37 of octane --------------------> ? of CO₂ gas

$= \frac{2,664.37 \times 704}{228.46} \\\\= 8,210.3 \ g\\\\= 8.21 \ kg$

Thus, the mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

Learn more about combustion of organic compounds here: brainly.com/question/13272422

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A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

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The production of iron and carbon dioxide from iron(iii) oxide and carbon monoxide is an exothermic reaction. which expression c

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Missing question:
A. [3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s)] / 2
B. 3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s) 
C. 26.3 kJ/1 mol Fe2O3 (s) / 3.40 mol Fe2O3 (s) 
D. 26.3 kJ/1 mol Fe2O3 (s) – 3.40 mol Fe2O3 (s).
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If 12.1 kilograms of al2o3(s), 60.4 kilograms of naoh(l), and 60.4 kilograms of hf(g) react completely, how many kilograms of cr

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Answer is: 7,826 kg of cryolite.
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n(Al₂O₃) = m(Al₂O₃) ÷ M(Al₂O₃).
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3 years ago

Yeast and other organisms can convert glucose (C6H12O6) to ethanol (CH3CH2OH) by a process called alchoholic fermentation. The n

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Answer:

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Explanation:

The balanced chemical equation is:

C₆H₁₂O₆ ⇒ 2 C₂H₅OH (l) + 2 CO₂ (g)

Now we are asked to calculate the mass of glucose required to produce 2.25 L CO₂ at 1atm and 295 K.

From the ideal gas law we can determine the number of moles that the 2.25 L represent.

From there we will use the stoichiometry of the reaction to determine the moles of glucose which knowing the molar mass can be converted to mass.

PV = nRT ⇒ n = PV/RT

n= 1 atm x 2.25 L / ( 0.08205 Latm/kmol x 295 K ) =0.093 mol CO₂

Moles glucose required:

0.093 mol CO₂ x ( 1 mol C₆H₁₂O₆ / 2 mol CO₂ ) = 0.046 mol C₆H₁₂O₆

The molar mass of glucose is 180.16 g/mol, then the mass required is

0.046 mol x 180.16 g/mol = 8.37 g

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2 years ago