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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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Calculate the grams of so2 gas present at stp in a 5. 9 l container. (atomic weights: s=32, o=16).

Solnce55 [7]

The 5.9 L container at STP contains 16.83 g of SO2.

To begin, we must determine how many moles of SO2 are contained in the 5.9 L container.

At standard temperature and pressure (STP), one mole of SO2 equates to 22.4 liters.

5.9 L = 5.9 / 22.45.9 L = 0.26 moles of SO2 as a result.

The result is that 0.263 moles of SO2 are present in the container.

Following that, we shall determine the mass of 0.263 moles of SO2.

The following are ways to do this:

The mass of SO2 is 32 + (16 2) = 32 + 32= 64 g/mol,

where SO2 Mole = 0.263 Molar.

Molar + Molar Mass Equals Mass

SO2 mass is 16.83 g, or 0.263 of a kilogram.

Therefore, 16.83 g of SO2 are present in the 5.9 L container at STP.

to know more about mass, visit to:-

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1 year ago

I need help with this :/

uranmaximum [27]

Ok please tell me what exactly you need help with

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3 years ago

The atomic number of arsenic is 33 what is the electron configuration

rjkz [21]

It is really easy if you look up an electron configuration table (one that looks like a periodic table) and then just go down the rows left to right, top to bottom, and just stop when you get to where the element is on the table.

$1s^2 2s^2 2p^{6} 3s^2 3p^6 4s^2 3d^1^0 4p^3$

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3 years ago

To begin the experiment, 1.11g of methane CH4is burned in a bomb calorimeter containing 1000 grams of water. The initial tempera

navik [9.2K]

1. CH4 + 2O2 → CO2 + 2H2O
2. 10.80 degrees celsius
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3 years ago

gulaghasi [49]

Taking into account the change of units, the size of the tank is 54.88 liters.

In first place, the rule of three is a tool used to quickly solve problems involving a proportional relationship between two variables.

The rule of three is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other), the direct rule of three should be applied.

The rule of three is applied as follows, with A, B and C being known values and D being an unknown value: if A maintains a certain relationship with B, and it is known that C and D have the same relationship, it is possible to calculate the unknown value D using the following expression:

$D=\frac{CxB}{A}$

In this case, the rule of three can be applied as follows: if 1 gallon is equal to 3.785 liters, 14.5 gallons are equal to how many liters?

$D=\frac{14.5 gallonsx3.785 liters}{1 gallon}$

Solving:

D= 54.88 liters

In summary, the size of the tank is 54.88 liters.

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2 years ago

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