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1 year ago

How many grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline?

1 answer:

5 0

Grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline is 45.5g

Octane is a hydrocarbon which burns in gasoline in presence of oxygen according to the given balanced chemical equation,

2C₈H₁₈ + 25O₂------> 16CO₂ + 18H₂0

Molar mass of octane = 114.23g/mol

Molar mass of Oxygen = 32g/mol

According to the stiochiometry of the balanced equation the mole ratio of Octane and Oxygen is 2:25

2 mole of octane needs 25 mole of oxygen

1 mole of octane needs 12.5 moleof oxygen

114.23g of octane needs 400g of oxygen

13g of octane needs 45.5g of oxygen

Mass of oxygen needed =45.5g

Hence, the Mass of oxygen needed is 45.5g for the combustion of octane in gasoline.

Learn more about Octane here, brainly.com/question/21268869

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Magnesium chloride is a salt formed with ionic bonds between one magnesium ion and two chloride ions. Magnesium has two electron

katrin [286]

Answer:

The magnesium atom loses 2 electron to the 2 atoms of chlorine. The 7 valency electrons of each chlorine atom will now be 8 to attain stable configuration. The final compound is written as MgCl2.

Explanation:

Ionic compounds are compound formed from the transfer of electron(s). One atom of the element loses electron(s) while the other atom gains electron(s).

The compound Magnesium chloride is an ionic compound . The bond between an atom of magnesium and 2 atoms of chlorine is an ionic bonding.

The valency electron of magnesium is 2 electron , for the atom of magnesium to attain octet rule, it will easily lose it 2 electrons to the chlorine atoms.

The chlorine atom on the other hand has 7 valency electrons, to attain octet configuration it will most likely gain 1 electron to become stable.

The magnesium atom loses 2 electron to the 2 atoms of chlorine. The 7 valency electrons of each chlorine atom will now be 8 to attain stable configuration. The final compound is written as MgCl2.

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A 2.1−mL volume of seawater contains about 4.0 × 10−10 g of gold. The total volume of ocean water is about 1.5 × 1021 L. Calcula

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Total worth of gold in the ocean = $5,840,000,000,000,000

Explanation:

As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.

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