Question

1. kclo₃ → kcl + o₂
2. na + h₂o → naoh + h₂

2na + 2h₂o → 2naoh + h₂

1. caco₃ → cao + co₂
2. pb(no₃)₂ + ki → pbi₂ + kno₃

Explanation:

Step1: Balance the first equation

We have $2KClO_3
ightarrow 2KCl + 3O_2$. We balance the oxygen - atoms first. Since there are 6 oxygen - atoms in $2KClO_3$, we need 3 moles of $O_2$. Then we balance the potassium and chlorine atoms which are already balanced with 2 moles of $KCl$.

Step2: Balance the second equation

For $2Na+2H_2O
ightarrow 2NaOH + H_2$. First, balance the hydrogen - atoms. There are 4 hydrogen - atoms on the left - hand side in $2H_2O$, so we need 2 moles of $NaOH$ and 1 mole of $H_2$. Then the sodium and oxygen atoms are also balanced.

Step3: Balance the third equation

The equation $CaCO_3
ightarrow CaO+CO_2$ is already balanced. There is 1 calcium, 1 carbon, and 3 oxygen atoms on both sides.

Step4: Balance the fourth equation

For $Pb(NO_3)_2 + 2KI
ightarrow PbI_2+2KNO_3$. We balance the nitrate and iodine atoms. There are 2 nitrate groups in $Pb(NO_3)_2$, so we need 2 moles of $KNO_3$. And there are 2 iodine atoms in $PbI_2$, so we need 2 moles of $KI$.

Answer:

1. $2KClO_3

ightarrow 2KCl + 3O_2$

1. $2Na+2H_2O

ightarrow 2NaOH + H_2$

1. $CaCO_3

ightarrow CaO+CO_2$

1. $Pb(NO_3)_2 + 2KI

ightarrow PbI_2+2KNO_3$