barium chlorate (ba(clo₃)₂) breaks down to form barium chloride and oxygen. what is the balanced equation…
barium chlorate (ba(clo₃)₂) breaks down to form barium chloride and oxygen. what is the balanced equation for this reaction?\no ba(clo₃)₂(s) → bacl₂(s) + o₂(g)\no ba(clo₃)₂(s) → bacl₂(s) + 3o₂(g)\no 2ba(clo₃)₂(s) → bacl(s) + 6o₂(g)\no ba(clo)₂(s) → bacl₂(s) + o₂(g)\ndone
Answer
Explanation:
Step1: Analyze reactant and products
The reactant is barium chlorate $\text{Ba(ClO}_3)_2$ and products are barium chloride $\text{BaCl}_2$ and oxygen $\text{O}_2$.
Step2: Balance oxygen atoms
In $\text{Ba(ClO}_3)_2$, there are 6 oxygen atoms. On the product - side, in $\text{O}_2$, we need to adjust the coefficient of $\text{O}_2$ to balance oxygen. If we put a coefficient of 3 in front of $\text{O}_2$, we have 6 oxygen atoms on the product - side as well.
Step3: Check other atoms
The barium and chlorine atoms are already balanced when we have $\text{Ba(ClO}_3)_2(s)\rightarrow\text{BaCl}_2(s)+3\text{O}_2(g)$. Barium has 1 atom on both sides and chlorine has 2 atoms on both sides.
Answer:
$\text{Ba(ClO}_3)_2(s)\rightarrow\text{BaCl}_2(s)+3\text{O}_2(g)$ (Second option)