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Reil [10]
2 years ago
14

Calculate the mass of oxygen produced from the decomposition of 75.0 g of potassium chlorate?

Chemistry
2 answers:
Dmitry [639]2 years ago
4 0
The balanced reaction is:

<span>2KClO3 = 2KCl + 3O2

</span>

We are given the amount of potassium chlorate. This will be our starting point.

 

75.0 g KClO3 (1 mol KClO3 /122.55 g KClO3) (3 mol O2/2 mol <span>KClO3</span>) ( 32 g O2 / 1mol O2) = 58.75 g O2

 

Thus, the answer is 28.88g H2O.

Elis [28]2 years ago
4 0

Answer:

29.4 g

Explanation:

Let's consider the decomposition of potassium chlorate.

KClO₃ → KCl + 1.5 O₂

The molar mass of KClO₃ is 122.55 g/mol. The moles of KClO₃ in 75.0 g are:

75.0 g × (1 mol/ 122.55 g) = 0.612 mol

The molar ratio of KClO₃ to O₂ is 1:1.5. The moles of O₂ are 1.5 × 0.612 mol = 0.918 mol

The molar mass of O₂ is 32.00 g/mol. The mass of O₂ corresponding to 0.918 moles is:

0.918 mol × (32.00 g/mol) = 29.4 g

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Answer:
             3.29 × 10²⁸ Atoms

Solution:
              Moles and Number of atoms are related to each other as,

                           Moles  =  Number of Atoms / 6.022 × 10²³   ------- (1)

Data Given;
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                  Number of Atoms  =  ?

Solving eq. 1 for Number of Atoms,

                        Number of Atoms  =  Moles  × 6.022 × 10²³

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Number of Atoms  =  3.29 × 10²⁸ Atoms
5 0
2 years ago
How did Mendeleev feel about Russian science education when he came back from Germany
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A container has a volume of 2.79 L and a pressure of 5.97 atm. If the pressure changes to 1460 mm Hg, what is the container’s ne
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Answer:

8.68 L is the new volume

Explanation:

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P_{1}V_{1}=P_{2}V_{2}

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V_{1} = first volume

V_{2} = second volume

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...maintain 3 significant figures in calculation, and round as needed...

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Hope this helps :)

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