Question

balancing chemical equations
balance the equations below:

1. _ n₂ + _ h₂ → ___ nh₃
2. _ kclo₃ → _ kcl + ___ o₂
3. _ nacl + _ f₂ → _ naf + _ cl₂
4. _ h₂ + _ o₂ → ___ h₂o
5. _ pb(oh)₂ + _ hcl → _ h₂o + _ pbcl₂
6. _ ch₄ + _ o₂ → _ co₂ + _ h₂o
7. _ p + _ o₂ → ___ p₂o₅
8. _ ag₂o → _ ag + ___ o₂
9. _ s₈ + _ o₂ → ___ so₃
10. _ k + _ mgbr₂ → _ kbr + _ mg

Explanation:

Step1: Balance nitrogen in $N_2 + H_2

ightarrow NH_3$
We have 2 nitrogen atoms on the left - hand side. To balance nitrogen, we put a 2 in front of $NH_3$. Then we balance hydrogen. We need 3 $H_2$ to get 6 hydrogen atoms on the left - hand side to match the 6 in $2NH_3$. So the balanced equation is $N_2+3H_2
ightarrow 2NH_3$.

Step2: Balance oxygen in $KClO_3

ightarrow KCl + O_2$
The least - common multiple of 3 (from $KClO_3$) and 2 (from $O_2$) for oxygen atoms is 6. So we put a 2 in front of $KClO_3$ and a 3 in front of $O_2$. Then we balance potassium and chlorine by putting a 2 in front of $KCl$. The balanced equation is $2KClO_3
ightarrow 2KCl + 3O_2$.

Step3: Balance chlorine and fluorine in $NaCl+F_2

ightarrow NaF + Cl_2$
We put a 2 in front of $NaCl$ and $NaF$ to balance sodium and chlorine. Then we have 2 fluorine atoms on the right - hand side, so we don't need to change the coefficient of $F_2$. The balanced equation is $2NaCl+F_2
ightarrow 2NaF + Cl_2$.

Step4: Balance oxygen in $H_2+O_2

ightarrow H_2O$
We put a 2 in front of $H_2O$ to balance oxygen. Then we balance hydrogen by putting a 2 in front of $H_2$. The balanced equation is $2H_2+O_2
ightarrow 2H_2O$.

Step5: Balance lead, chlorine, oxygen and hydrogen in $Pb(OH)_2+HCl

ightarrow H_2O+PbCl_2$
We put a 2 in front of $HCl$ and $H_2O$ to balance chlorine, hydrogen and oxygen. The balanced equation is $Pb(OH)_2 + 2HCl
ightarrow 2H_2O+PbCl_2$.

Step6: Balance carbon, hydrogen and oxygen in $CH_4+O_2

ightarrow CO_2+H_2O$
We first balance carbon by having 1 $CO_2$ for 1 $CH_4$. Then we balance hydrogen. Since there are 4 hydrogen atoms in $CH_4$, we put a 2 in front of $H_2O$. Then we balance oxygen. We need 2 $O_2$ to get 4 oxygen atoms on the left - hand side. The balanced equation is $CH_4+2O_2
ightarrow CO_2 + 2H_2O$.

Step7: Balance phosphorus and oxygen in $P+O_2

ightarrow P_2O_5$
The least - common multiple of 1 (from $P$) and 2 (from $P_2$) for phosphorus is 2, and for oxygen, the least - common multiple of 2 (from $O_2$) and 5 (from $P_2O_5$) is 10. We put a 4 in front of $P$ and a 5 in front of $O_2$ and a 2 in front of $P_2O_5$. The balanced equation is $4P+5O_2
ightarrow 2P_2O_5$.

Step8: Balance silver and oxygen in $Ag_2O

ightarrow Ag+O_2$
We put a 2 in front of $Ag_2O$ to balance oxygen, and then a 4 in front of $Ag$ to balance silver. The balanced equation is $2Ag_2O
ightarrow 4Ag+O_2$.

Step9: Balance sulfur and oxygen in $S_8+O_2

ightarrow SO_3$
We put an 8 in front of $SO_3$ to balance sulfur. Then we need 12 $O_2$ to get 24 oxygen atoms on the left - hand side. The balanced equation is $S_8+12O_2
ightarrow 8SO_3$.

Step10: Balance potassium, magnesium and bromine in $K+MgBr_2

ightarrow KBr+Mg$
We put a 2 in front of $K$ and $KBr$ to balance potassium and bromine. The balanced equation is $2K+MgBr_2
ightarrow 2KBr+Mg$.

Answer:

1. $1N_2 + 3H_2

ightarrow 2NH_3$

1. $2KClO_3

ightarrow 2KCl + 3O_2$

1. $2NaCl+F_2

ightarrow 2NaF + Cl_2$

1. $2H_2+O_2

ightarrow 2H_2O$

1. $Pb(OH)_2 + 2HCl

ightarrow 2H_2O+PbCl_2$

1. $CH_4+2O_2

ightarrow CO_2 + 2H_2O$

1. $4P+5O_2

ightarrow 2P_2O_5$

1. $2Ag_2O

ightarrow 4Ag+O_2$

1. $S_8+12O_2

ightarrow 8SO_3$

1. $2K+MgBr_2

ightarrow 2KBr+Mg$