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Reil [10]
3 years ago
14

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

Chemistry
2 answers:
Dmitry [639]3 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]3 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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Luda [366]

Answer:

* pH=4.37

* The solution is acidic since the pH is below 7.

Explanation:

Hello,

In this case, we can mathematically define the pH by:

pH=-log([H_3O^+])

Thus, for the given hydronium concentration we simply compute the pH:

pH=-log(4.3x10^{-5})=4.37

Thereby, we conclude the solution is acidic due to the fact that the pH is below 7 which is the neutral point and above it the solutions are basic.

Regards.

3 0
3 years ago
The smallest unit which maintains the physical properties of a compound is a(n _____. atom element molecule alloythe smallest un
LuckyWell [14K]
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5 0
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Make a prediction about the relationship between electrons and molecular shapes
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3 years ago
How much water (H2O ) would form if 4.04 g of hydrogen (H2) reacted with 31.98 g of oxygen (O2 )?
Norma-Jean [14]

Answer:

Mass = 36 g

Explanation:

Given data:

Mass of water formed = ?

Mass of hydrogen = 4.04 g

Mass of oxygen = 31.98 g

Solution:

Chemical equation:

2H₂ + O₂   →   2H₂O

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 4.04 g/ 2 g/mol

Number of moles = 2.02 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 31.98 g/ 32 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of water with hydrogen and oxygen.

                O₂         :         H₂O

                 1           :           2

                H₂         :         H₂O

                 2          :          2

               2.02       :      2.02

Number of moles of water formed by oxygen are less thus oxygen will limiting reactant.

Mass of water:

Mass = number of moles × molar mass

Mass = 2 mol × 18 g/mol

Mass = 36 g

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