Question

now that we know the \types of reactions\ it is important for us to also \balance\ those equations. a few of the coefficients are there to help get you started!
\boxed{} kclo_3 \
ightarrow \boxed{} kcl + \underline{\\ 3\\ } o_2
\underline{\\ 1\\ } c_6h_{12}o_6 + \underline{\\ 6\\ } o_2 \
ightarrow \boxed{} co_2 + \underline{\\ 6\\ } h_2o

Explanation:

First Equation: Balancing $\boldsymbol{KClO_3

ightarrow KCl + O_2}$

Step1: Balance Oxygen atoms

On the right, $O_2$ has 2 O atoms, and on the left, $KClO_3$ has 3 O atoms. The least common multiple of 2 and 3 is 6. So, we put a coefficient of 3 in front of $O_2$ (already given) and a coefficient of 2 in front of $KClO_3$ to get 6 O atoms on both sides:
$2 KClO_3
ightarrow \square KCl + 3 O_2$

Step2: Balance K and Cl atoms

Now, with 2 $KClO_3$, we have 2 K and 2 Cl atoms on the left. So, we put a coefficient of 2 in front of $KCl$ to balance K and Cl:
$2 KClO_3
ightarrow 2 KCl + 3 O_2$

Second Equation: Balancing $\boldsymbol{C_6H_{12}O_6 + O_2

ightarrow CO_2 + H_2O}$

Step1: Balance C atoms

On the left, $C_6H_{12}O_6$ has 6 C atoms. So, we put a coefficient of 6 in front of $CO_2$ to balance C:
$1 C_6H_{12}O_6 + 6 O_2
ightarrow 6 CO_2 + 6 H_2O$

Final Answers:

• For $KClO_3

ightarrow KCl + O_2$: The coefficients are $\boldsymbol{2}$ (for $KClO_3$) and $\boldsymbol{2}$ (for $KCl$).

• For $C_6H_{12}O_6 + O_2

ightarrow CO_2 + H_2O$: The coefficient for $CO_2$ is $\boldsymbol{6}$.

Answer:

Step1: Balance C atoms

On the left, $C_6H_{12}O_6$ has 6 C atoms. So, we put a coefficient of 6 in front of $CO_2$ to balance C:
$1 C_6H_{12}O_6 + 6 O_2
ightarrow 6 CO_2 + 6 H_2O$

Final Answers:

• For $KClO_3

ightarrow KCl + O_2$: The coefficients are $\boldsymbol{2}$ (for $KClO_3$) and $\boldsymbol{2}$ (for $KCl$).

• For $C_6H_{12}O_6 + O_2

ightarrow CO_2 + H_2O$: The coefficient for $CO_2$ is $\boldsymbol{6}$.