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2 years ago

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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Suppose 110.0 mL110.0 mL of hydrogen gas at STP combines with a stoichiometric amount of fluorine gas and the resulting hydrogen

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Suppose 110.0 mL of hydrogen gas at STP combines with a stoichiometric amount of fluorine gas and the resulting hydrogen fluoride dissolves in water to form 150.0 mL of an aqueous solution. 0.032 M is the concentration of the resulting hydrofluoric acid.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

Now write the balanced chemical equation

H₂ + F₂ → 2HF

<h3>What is Ideal Gas ?</h3>

An ideal gas is a gas that obey gas laws at all temperature and pressure conditions. It have velocity and mass but do not have volume. Ideal gas is also called perfect gas. Ideal gas is a hypothetical gas.

It is expressed as:

PV = nRT

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V = Volume

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T = temperature

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P = 1 atm [At STP]

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T = 273 K [At STP]

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PV = nRT

1 atm × 0.11 L = n × 0.0821 L.atm/ K. mol × 273 K

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To calculate the concentration of resulting solution use the expression

$C = \frac{n}{V}$

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