__ KClO₃ → __ KCl + __ O₂
Left Side:
1 K
1 Cl
3 O
Right Side:
1 K
1 Cl
2 O
Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.
This gives us 2KClO₃ → __ KCl + 3O₂.
However, this equation is still not balanced.
Left Side:
2 K
2 Cl
6 O
Right Side:
1 K
1 Cl
6 O
In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.
2KClO₃ → 2KCl + 3O₂
Answer:
la particula que queda es h2o
Explanation:
Answer:
Entropy increases
Explanation:
Entropy (S) is a measure of the degree of disorder. For a given substance - say water - across phases the following is true ...
S(ice) < S(water) << S(steam)
For a chemical process, entropy changes can be related to increasing or decreasing molar volumes of gas from reactant side of equation to product side of equation. That is ...
if molar volumes of gas increase, then entropy increases, and
if molar volumes of gas decrease, then entropy decreases.
For the reaction 2KClO₃(s) => 2KCl(s) + 3O₂(g)
molar volumes of gas => 0Vm* 0Vm 3Vm
*molar volumes (Vm) apply only to gas phase substances. Solids and liquids do not have molar volume.
Since the reaction produces 3 molar volumes of O₂(g) product vs 0 molar volumes of reactant, then the reaction is showing an increase in molar volumes of gas phase substances and its entropy is therefore increasing.
