Question

4.90g of potassium chlorate when heated produced 1.92g of oxygen & the residue illustrate the law of conservation of mass.

Answer 1

Answer : This reaction follow 'Law of conservation of mass'.

Solution : Given,

Mass of potassium chlorate = 4.90 g

Mass of oxygen = 1.92 g

Molar mass of oxygen = 32 g/mole

Molar mass of potassium chlorate = 122.55 g/mole

Molar mass of potassium chloride = 74.55 g/mole

The balanced chemical reaction is,

$2KClO_3 \overset{\Delta}{\rightarrow}2KCl+3O_2$

Step 1 : First we have to calculate the moles of potassium chlorate.

$\text{ Moles of } KClO_3=\frac{\text{ mass of } KClO_3}{\text{ molar mass of } KClO_3}$ = $\frac{4.9g}{122.55g/mole}=0.0399moles$

Step 2 : Now we have to calculate mass of $O_2$.

From the balanced chemical reaction, we conclude that

2 moles of potassium chlorate produced 3 moles of oxygen

and 0.0399 moles of potassium chlorate produced $[\frac{3moles}{2moles}\times 0.0399moles]$ moles of oxygen

The moles of oxygen produced = 0.05985 moles

The mass of oxygen produced = Number of moles of oxygen × Molar mass of oxygen = 0.05985 moles × 32 g/mole = 1.9152 g

Step 3 : Now we have to calculate the mass of KCl.

From the balanced chemical reaction, we conclude that

2 moles of potassium chlorate produced 2 moles of KCl

and 0.0399 moles of potassium chlorate produced $[\frac{2moles}{2moles}\times 0.0399moles]$ moles of KCl

The moles of KCl produced = 0.0399 moles

The mass of KCl produced = Number of moles of KCl × Molar mass of KCl = 0.0399 moles × 74.55 g/mole = 2.974 g

According to the 'Law of conservation of mass', the mass of reactant side is equal to the mass of product side.

The mass of reactant side = mass of potassium chlorate = 4.90 g

The mass of product side = mass of KCl + mass of oxygen = (1.9152 + 2.974)g = 4.889 g ≈ 4.9 g

Thus, we conclude that the mass of reactant is equal to the mass of product.

Hence, this reaction follow the 'Law of conservation of mass'.

Answer 2

Answer:

Answer : This reaction follow 'Law of conservation of mass'.

Solution : Given,

Mass of potassium chlorate = 4.90 g

Mass of oxygen = 1.92 g

Molar mass of oxygen = 32 g/mole

Molar mass of potassium chlorate = 122.55 g/mole

Molar mass of potassium chloride = 74.55 g/mole

The balanced chemical reaction is,

2KClO_3 \overset{\Delta}{\rightarrow}2KCl+3O_22KClO

3

→

Δ

2KCl+3O

2

Step 1 : First we have to calculate the moles of potassium chlorate.

\text{ Moles of } KClO_3=\frac{\text{ mass of } KClO_3}{\text{ molar mass of } KClO_3} Moles of KClO

3

=

molar mass of KClO

3

mass of KClO

3

= \frac{4.9g}{122.55g/mole}=0.0399moles

122.55g/mole

4.9g

=0.0399moles

Step 2 : Now we have to calculate mass of O_2O

2

.

From the balanced chemical reaction, we conclude that

2 moles of potassium chlorate produced 3 moles of oxygen

and 0.0399 moles of potassium chlorate produced [\frac{3moles}{2moles}\times 0.0399moles][

2moles

3moles

×0.0399moles] moles of oxygen

The moles of oxygen produced = 0.05985 moles

The mass of oxygen produced = Number of moles of oxygen × Molar mass of oxygen = 0.05985 moles × 32 g/mole = 1.9152 g

Step 3 : Now we have to calculate the mass of KCl.

From the balanced chemical reaction, we conclude that

2 moles of potassium chlorate produced 2 moles of KCl

and 0.0399 moles of potassium chlorate produced [\frac{2moles}{2moles}\times 0.0399moles][

2moles

2moles

×0.0399moles] moles of KCl

The moles of KCl produced = 0.0399 moles

The mass of KCl produced = Number of moles of KCl × Molar mass of KCl = 0.0399 moles × 74.55 g/mole = 2.974 g

According to the 'Law of conservation of mass', the mass of reactant side is equal to the mass of product side.

The mass of reactant side = mass of potassium chlorate = 4.90 g

The mass of product side = mass of KCl + mass of oxygen = (1.9152 + 2.974)g = 4.889 g ≈ 4.9 g

Thus, we conclude that the mass of reactant is equal to the mass of product.

Hence, this reaction follow the 'Law of conservation of mass'.

Explanation:

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