Question

steps for balancing equations
hgo→hg + __o₂
hg - 1
o - 1
hg - 1
o - 2
2al+3cucl₂→3cu + 2alcl₃
al - 1
al - 1
cu - 1
cu - 1
cl - 2
cl - 3
partner practice: n₂+h₂→__nh₃
n - 2
n - 1
h - 2
h - 3
partner practice: kclo₃→kcl+__o₂
k - 1
cl - 1
o - 3
k - 1
cl - 1
o - 2
partner practice: kbr+cl₂→kcl+br₂
k - 1
br - 1
cl - 2
k - 1
cl - 1
br - 2
partner practice: co+o₂→__co₂
c - 1
c - 1
o - 2
o - 2
solo practice: 4c+s₈→4cs₂
c - 1
c - 1
s - 8
s - 2
solo practice: p₄+br₂→__pbr₃
p - 4
p - 1
br - 2
br - 3
solo practice: na₃p+caf₂→naf+ca₃p₂
na - 1
na - 1
p - 1
p - 1
ca - 1
ca - 3
f - 2
f - 1

Explanation:

Step1: Balance the mercury and oxygen in $HgO

ightarrow Hg+O_2$
We have 1 oxygen atom on the left - hand side and 2 on the right - hand side. Multiply $HgO$ by 2 to get $2HgO
ightarrow 2Hg + O_2$.

Step2: Balance $N_2 + H_2

ightarrow NH_3$
There are 2 nitrogen atoms in $N_2$ and 1 in $NH_3$, and 2 hydrogen atoms in $H_2$ and 3 in $NH_3$. Multiply $N_2$ by 1, $H_2$ by 3 and $NH_3$ by 2 to get $N_2+3H_2
ightarrow 2NH_3$.

Step3: Balance $KClO_3

ightarrow KCl + O_2$
There are 3 oxygen atoms in $KClO_3$ and 2 in $O_2$. Multiply $KClO_3$ by 2, $KCl$ by 2 and $O_2$ by 3 to get $2KClO_3
ightarrow 2KCl + 3O_2$.

Step4: Balance $KBr+Cl_2

ightarrow KCl + Br_2$
There are 2 chlorine atoms in $Cl_2$ and 1 in $KCl$, and 2 bromine atoms in $Br_2$ and 1 in $KBr$. Multiply $KBr$ by 2, $KCl$ by 2 to get $2KBr+Cl_2
ightarrow 2KCl + Br_2$.

Step5: Balance $CO + O_2

ightarrow CO_2$
There are 2 oxygen atoms in $O_2$ and 1 in $CO$. Multiply $CO$ by 2 and $CO_2$ by 2 to get $2CO+O_2
ightarrow 2CO_2$.

Step6: Balance $C + S_8

ightarrow CS_2$
There are 8 sulfur atoms in $S_8$ and 2 in $CS_2$. Multiply $CS_2$ by 4 and $C$ by 4 to get $4C + S_8
ightarrow 4CS_2$.

Step7: Balance $P_4+Br_2

ightarrow PBr_3$
There are 4 phosphorus atoms in $P_4$ and 1 in $PBr_3$. Multiply $PBr_3$ by 4. There are 2 bromine atoms in $Br_2$ and 3 in $PBr_3$. Multiply $Br_2$ by 6 to get $P_4 + 6Br_2
ightarrow 4PBr_3$.

Step8: Balance $Na_3P+CaF_2

ightarrow NaF+Ca_3P_2$
There are 3 sodium atoms in $Na_3P$ and 1 in $NaF$, 1 phosphorus atom in $Na_3P$ and 2 in $Ca_3P_2$, 1 calcium atom in $CaF_2$ and 3 in $Ca_3P_2$, 2 fluorine atoms in $CaF_2$ and 1 in $NaF$. Multiply $Na_3P$ by 2, $CaF_2$ by 3, $NaF$ by 6 to get $2Na_3P+3CaF_2
ightarrow 6NaF + Ca_3P_2$.

Answer:

$2HgO
ightarrow 2Hg + O_2$; $N_2+3H_2
ightarrow 2NH_3$; $2KClO_3
ightarrow 2KCl + 3O_2$; $2KBr+Cl_2
ightarrow 2KCl + Br_2$; $2CO+O_2
ightarrow 2CO_2$; $4C + S_8
ightarrow 4CS_2$; $P_4 + 6Br_2
ightarrow 4PBr_3$; $2Na_3P+3CaF_2
ightarrow 6NaF + Ca_3P_2$