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Sedaia [141]
3 years ago
12

The reaction that occurs in a Breathalyzer, a device used to determine the alcohol level in a person's bloodstream, is given bel

ow. If the rate of appearance of Cr2(SO4)3 is 1.24 mol/min at a particular moment, what is the rate of disappearance of C2H6O at that moment
Chemistry
1 answer:
Umnica [9.8K]3 years ago
4 0

Answer:

The rate of disappearance of C₂H₆O = 2.46 mol/min

Explanation:

The equation of the reaction is given below:

2 K₂Cr₂O₇ + 8 H₂SO₄ + 3 C₂H₆O → 2 Cr₂(SO₄)₃ + 2 K₂SO₄ + 11 H₂O

From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.

The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus,  the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.

Rate of disappearance of C₂H₆O = 1.64 mol/min of  Cr₂(SO₄)₃  * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃

Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O

Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min

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How many gallons of 80% antifreeze solution must be mixed with 60 gallons of 20% antifreeze to get a mixture that is 70% antifre
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Explanation:

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If 38.5 grams of potassium react with excess oxygen gas, how many grams of potassium oxide can be produced? 4K + O2 yields 2K2O
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Answer:

46.40 g.

Explanation:

  • It is a stichiometric problem.
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  • It is clear that 4.0 moles of K reacts with 1.0 mole of oxygen produces 2.0 moles of K₂O.
  • We should convert the mass of K (38.5 g) into moles using the relation:

<em>n = mass / molar mass,</em>

n = (38.5 g) / (39.098 g/mol) = 0.985 mole.

<em>Using cross multiplication:</em>

4.0 moles of K produces → 2.0 moles of K₂O, from the stichiometry.

0.985 mole of K produces → ??? moles of K₂O.

∴ The number of moles of K₂O produced = (0.985 mole) (2.0 mole) / (4.0 mole) = 0.4925 mole ≅ 0.5 mole.

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