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agasfer [191]
3 years ago
10

If this reaction produced 19.6 g of kcl, how much o2 was produced (in grams)?

Chemistry
1 answer:
stepladder [879]3 years ago
4 0
Answer is: mass of oxygen is 12,576 grams.
Missing chemical reaction: 2KClO₃ → 2KCl + 3O₂.
m(KCl) = 19,6 g.
n(KCl) = m(KCl) ÷ M(KCl).
n(KCl) = 19,6 g ÷ 74,55 g/mol.
n(KCl) = 0,262 mol.
From chemical reaction: n(KCl) : n(O₂) = 2 : 3.
n(O₂) = 3 · 0,262 mol ÷ 2.
n(O₂) = 0,393 mol.
m(O₂) = n(O₂) · M(O₂).
m(O₂) = 0,393 mol · 32 g/mol.
m(O₂) = 12,576 g.

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The state after the tests must guarantee the population the treatment of the water to reduce the levels of lead. The main pipes that contain lead pipes must be changed, as well as those parts of the service connections made of lead.

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A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

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