Question

balance the equations by adding whole number co

1. h₂ + o₂ → h₂o
2. n₂ + h₂ → nh₃
3. c₃h₈ + o₂ → co₂ + h₂o
4. fe + o₂ → fe₂o₃
5. al + hcl → alcl₃ + h₂
6. kclo₃ → kcl + o₂
7. na + h₂o → naoh + h₂

Explanation:

Step1: Balance hydrogen and oxygen in $H_2 + O_2

ightarrow H_2O$
For hydrogen, we have 2 on the left and 2 on the right in the un - balanced equation for hydrogen in $H_2O$. For oxygen, we have 2 on the left and 1 on the right. Multiply $H_2O$ by 2 and $H_2$ by 2 to get $2H_2+O_2 = 2H_2O$.

Step2: Balance nitrogen and hydrogen in $N_2 + H_2

ightarrow NH_3$
There are 2 nitrogen atoms on the left. So, we need 2 moles of $NH_3$ on the right. Then, to balance hydrogen, since we have 2 moles of $NH_3$ (6 hydrogen atoms), we multiply $H_2$ by 3. The balanced equation is $N_2 + 3H_2=2NH_3$.

Step3: Balance carbon, hydrogen and oxygen in $C_3H_8+O_2

ightarrow CO_2 + H_2O$
First, balance carbon. There are 3 carbon atoms in $C_3H_8$, so we need 3 moles of $CO_2$. There are 8 hydrogen atoms in $C_3H_8$, so we need 4 moles of $H_2O$. Then, for oxygen, we have $3\times2 + 4\times1=10$ oxygen atoms on the right - hand side. So, we multiply $O_2$ by 5. The balanced equation is $C_3H_8+5O_2 = 3CO_2+4H_2O$.

Step4: Balance iron and oxygen in $Fe + O_2

ightarrow Fe_2O_3$
To balance oxygen, the least common multiple of 2 and 3 (the number of oxygen atoms in $O_2$ and $Fe_2O_3$ respectively) is 6. Multiply $O_2$ by 3 and $Fe_2O_3$ by 2 to get 6 oxygen atoms on both sides. Then, to balance iron, since we have $2Fe_2O_3$ (4 iron atoms), we multiply $Fe$ by 4. The balanced equation is $4Fe+3O_2 = 2Fe_2O_3$.

Step5: Balance aluminum, hydrogen and chlorine in $Al + HCl

ightarrow AlCl_3+H_2$
First, balance chlorine. There are 3 chlorine atoms in $AlCl_3$. So, we multiply $HCl$ by 3. Then, to balance hydrogen, since we have 3 moles of $HCl$ (3 hydrogen atoms) and we want an even number of hydrogen atoms (because of $H_2$), we multiply all coefficients by 2. The balanced equation is $2Al + 6HCl=2AlCl_3+3H_2$.

Step6: Balance potassium, chlorine and oxygen in $KClO_3

ightarrow KCl + O_2$
The least common multiple of 3 and 2 (the number of oxygen atoms in $KClO_3$ and $O_2$ respectively) is 6. Multiply $KClO_3$ by 2 and $O_2$ by 3 to balance oxygen. Then, to balance potassium and chlorine, since we have 2 moles of $KClO_3$, we need 2 moles of $KCl$. The balanced equation is $2KClO_3=2KCl + 3O_2$.

Step7: Balance sodium, hydrogen and oxygen in $Na + H_2O

ightarrow NaOH + H_2$
First, balance hydrogen. Multiply $H_2O$ by 2 and $NaOH$ by 2 to get 4 hydrogen atoms on both sides. Then, to balance sodium, since we have 2 moles of $NaOH$, we multiply $Na$ by 2. The balanced equation is $2Na+2H_2O = 2NaOH+H_2$.

Answer:

1. $2H_2+O_2 = 2H_2O$
2. $N_2 + 3H_2=2NH_3$
3. $C_3H_8+5O_2 = 3CO_2+4H_2O$
4. $4Fe+3O_2 = 2Fe_2O_3$
5. $2Al + 6HCl=2AlCl_3+3H_2$
6. $2KClO_3=2KCl + 3O_2$
7. $2Na+2H_2O = 2NaOH+H_2$