Answer:
Explanation:
At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).
Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.
Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).
A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.
<u>a. C₂H₄:</u>
- C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)
Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.
The following analysis just shows that the other options are not right.
<u>b. C₂H₂:</u>
- 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)
The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.
<u>с. С₃Н₈</u>
- C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)
The mole ratio is 1 mol C₃H₈ : 5 mol O₂
<u>d. C₂H₆</u>
- 2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)
The mole ratio is 2 mol C₂H₆ : 7 mol O₂
Answer: - 0.75
Explanation:
For this question, you must remember that a fraction can be written of division.
So, -3/4 = -0.75, between -0.7 and -0.8 on the number line.
MgCl2 is a ionic compound because it is formed by a metal (Mg) and a non-metal (Cl)
The number of mole of nitrogen gas, N₂, needed to produce 150 g of ammonia, NH₃ is 4.41 moles
<h3>How to determine the mole of NH₃ produced </h3>
- Mass of NH₃ = 150 g
- Molar mass of NH₃ = 14 + (3×1) = 17 g/mol
Mole = mass /molar mass
Mole of NH₃ = 150 / 17
Mole of NH₃ = 8.82 moles
<h3>How to determine the mole of N₂ needed </h3>
Balanced equation
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
2 moles of NH₃ were produced by 1 mole of N₂.
Therefore,
8.82 moles of NH₃ will be produced by = 8.82 / 2 = 4.41 moles of N₂.
Thus, 4.41 moles of N₂ is needed for the reaction.
Learn more about stoichiometry:
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