Explain how to balance the chemical equation, KClO3 ⟶ KCl + O2 and include all steps

The balance of chemical equation : 2KClO₃ ⟶ 2KCl + 3O₂

Equalization (balance) of chemical reactions can be done using variables. Steps in equalizing the reaction equation:  

Reaction

1. Coefficient

KClO₃ ⟶ aKCl + bO₂

2. Equation

K ⇒ left =1, right=a ⇒a=1

Cl⇒ left =1, right=a ⇒a=1

O ⇒ left = 3, right = 2b ⇒3=2b, b = 1.5

The equation turns into :

KClO₃ ⟶ KCl + 1.5O₂

multiplied by 2

2KClO₃ ⟶ 2KCl + 3O₂

out of these options:

ability to burn paper

pure sodium reacting violently with water

the boiling point of water

helium does not tend to react with anything

i would say this is a physical property:

helium does not tend to react with anything

why? :

i said helium does not tend to react with anything because well, it reacts to ! but ability to burn paper, pure sodium reacting violently with water, & the boiling point of water are chemical or

physical properties: the properties of a substance that can be observed without changing the chemical makeup of the substance

chemical properties: properties of a substance relating to the chemical nature and reactivity of the substance

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%