Question

make the following equations on your desk
reactants start
products start
reactants final
products final
$\ce{h_{2} + o_{2} -> h_{2}o}$
$\ce{h_{2}o_{2} -> h_{2}o + o_{2}}$
$\ce{na + o_{2} -> na_{2}o}$
$\ce{n_{2} + h_{2} -> nh_{3}}$
$\ce{p_{4} + o_{2} -> p_{4}o_{10}}$
$\ce{fe + h_{2}o -> fe_{3}o_{4} + h_{2}}$
$\ce{c + h_{2} -> ch_{4}}$
$\ce{na_{2}so_{4} + cacl_{2} -> caso_{4} + nacl}$
$\ce{c_{2}h_{6} + o_{2} -> co_{2} + h_{2}o}$
$\ce{al_{2}o_{3} -> al + o_{2}}$

Answer:

To balance these chemical equations, we'll use the method of balancing by adjusting coefficients to ensure the number of each atom is the same on both sides of the equation.

1. $\boldsymbol{H_2 + O_2

ightarrow H_2O}$

Step 1: Count atoms

• Reactants: $H = 2$, $O = 2$
• Products: $H = 2$, $O = 1$

Step 2: Balance O

Multiply $H_2O$ by 2: $H_2 + O_2
ightarrow 2H_2O$
Now, $H$ on products: $4$

Step 3: Balance H

Multiply $H_2$ by 2: $2H_2 + O_2
ightarrow 2H_2O$
Check: $H = 4$, $O = 2$ on both sides.
Balanced: $\boldsymbol{2H_2 + O_2 = 2H_2O}$

2. $\boldsymbol{H_2O_2

ightarrow H_2O + O_2}$

Step 1: Count atoms

• Reactants: $H = 2$, $O = 2$
• Products: $H = 2$, $O = 3$ (1 in $H_2O$, 2 in $O_2$)

Step 2: Balance O

Multiply $H_2O_2$ by 2, $H_2O$ by 2: $2H_2O_2
ightarrow 2H_2O + O_2$
Check: $H = 4$, $O = 4$ (2×2 in reactants; 2×1 + 2 in products)
Balanced: $\boldsymbol{2H_2O_2 = 2H_2O + O_2}$

3. $\boldsymbol{Na + O_2

ightarrow Na_2O}$

Step 1: Count atoms

• Reactants: $Na = 1$, $O = 2$
• Products: $Na = 2$, $O = 1$

Step 2: Balance O

Multiply $Na_2O$ by 2: $Na + O_2
ightarrow 2Na_2O$
Now, $Na$: 4, $O$: 2

Step 3: Balance Na

Multiply $Na$ by 4: $4Na + O_2
ightarrow 2Na_2O$
Check: $Na = 4$, $O = 2$
Balanced: $\boldsymbol{4Na + O_2 = 2Na_2O}$

4. $\boldsymbol{N_2 + H_2

ightarrow NH_3}$

Step 1: Count atoms

• Reactants: $N = 2$, $H = 2$
• Products: $N = 1$, $H = 3$

Step 2: Balance N

Multiply $NH_3$ by 2: $N_2 + H_2
ightarrow 2NH_3$
Now, $H$: 6

Step 3: Balance H

Multiply $H_2$ by 3: $N_2 + 3H_2
ightarrow 2NH_3$
Check: $N = 2$, $H = 6$
Balanced: $\boldsymbol{N_2 + 3H_2 = 2NH_3}$

5. $\boldsymbol{P_4 + O_2

ightarrow P_4O_{10}}$

Step 1: Count atoms

• Reactants: $P = 4$, $O = 2$
• Products: $P = 4$, $O = 10$

Step 2: Balance O

Multiply $O_2$ by 5: $P_4 + 5O_2
ightarrow P_4O_{10}$
Check: $P = 4$, $O = 10$
Balanced: $\boldsymbol{P_4 + 5O_2 = P_4O_{10}}$

6. $\boldsymbol{Fe + H_2O

ightarrow Fe_3O_4 + H_2}$

Step 1: Count atoms

• Reactants: $Fe = 1$, $H = 2$, $O = 1$
• Products: $Fe = 3$, $H = 2$, $O = 4$

Step 2: Balance Fe

Multiply $Fe$ by 3: $3Fe + H_2O
ightarrow Fe_3O_4 + H_2$
Now, $O$: 4 (in $Fe_3O_4$)

Step 3: Balance O

Multiply $H_2O$ by 4: $3Fe + 4H_2O
ightarrow Fe_3O_4 + H_2$
Now, $H$: 8 (in $4H_2O$)

Step 4: Balance H

Multiply $H_2$ by 4: $3Fe + 4H_2O
ightarrow Fe_3O_4 + 4H_2$
Check: $Fe = 3$, $H = 8$, $O = 4$
Balanced: $\boldsymbol{3Fe + 4H_2O = Fe_3O_4 + 4H_2}$

7. $\boldsymbol{C + H_2

ightarrow CH_4}$

Step 1: Count atoms

• Reactants: $C = 1$, $H = 2$
• Products: $C = 1$, $H = 4$

Step 2: Balance H

Multiply $H_2$ by 2: $C + 2H_2
ightarrow CH_4$
Check: $C = 1$, $H = 4$
Balanced: $\boldsymbol{C + 2H_2 = CH_4}$

8. $\boldsymbol{Na_2SO_4 + CaCl_2

ightarrow CaSO_4 + NaCl}$

Step 1: Count atoms

• Reactants: $Na = 2$, $S = 1$, $O = 4$, $Ca = 1$, $Cl = 2$
• Products: $Na = 1$, $S = 1$, $O = 4$, $Ca = 1$, $Cl = 1$

Step 2: Balance Na and Cl

Multiply $NaCl$ by 2: $Na_2SO_4 + CaCl_2
ightarrow CaSO_4 + 2NaCl$
Check: $Na = 2$, $Cl = 2$, others balanced.
Balanced: $\boldsymbol{Na_2SO_4 + CaCl_2 = CaSO_4 + 2NaCl}$

9. $\boldsymbol{C_2H_6 + O_2

ightarrow CO_2 + H_2O}$

Step 1: Count atoms

• Reactants: $C = 2$, $H = 6$, $O = 2$
• Products: $C = 1$, $H = 2$, $O = 3$ (2 in $CO_2$, 1 in $H_2O$)

Step 2: Balance C

Multiply $CO_2$ by 2: $C_2H_6 + O_2
ightarrow 2CO_2 + H_2O$
Now, $C = 2$, $H = 6$, $O = 5$ (4 + 1)

Step 3: Balance H

Multiply $H_2O$ by 3: $C_2H_6 + O_2
ightarrow 2CO_2 + 3H_2O$
Now, $H = 6$, $O = 7$ (4 + 3)

Step 4: Balance O

Multiply $O_2$ by $\frac{7}{2}$: $C_2H_6 + \frac{7}{2}O_2
ightarrow 2CO_2 + 3H_2O$
Multiply all by 2 to eliminate fraction: $2C_2H_6 + 7O_2
ightarrow 4CO_2 + 6H_2O$
Check: $C = 4$, $H = 12$, $O = 14$ (7×2; 4×2 + 6×1)
Balanced: $\boldsymbol{2C_2H_6 + 7O_2 = 4CO_2 + 6H_2O}$

10. $\boldsymbol{Al_2O_3

ightarrow Al + O_2}$

Step 1: Count atoms

• Reactants: $Al = 2$, $O = 3$
• Products: $Al = 1$, $O = 2$

Step 2: Balance Al

Multiply $Al$ by 2: $Al_2O_3
ightarrow 2Al + O_2$
Now, $Al = 2$, $O = 3$ (reactants), $O = 2$ (products)

Step 3: Balance O

Find LCM of 3 and 2 (6). Multiply $Al_2O_3$ by 2, $O_2$ by 3: $2Al_2O_3
ightarrow 4Al + 3O_2$
Check: $Al = 4$, $O = 6$ (2×3; 3×2)
Balanced: $\boldsymbol{2Al_2O_3 = 4Al + 3O_2}$

Final Balanced Equations:

1. $2H_2 + O_2 = 2H_2O$
2. $2H_2O_2 = 2H_2O + O_2$
3. $4Na + O_2 = 2Na_2O$
4. $N_2 + 3H_2 = 2NH_3$
5. $P_4 + 5O_2 = P_4O_{10}$
6. $3Fe + 4H_2O = Fe_3O_4 + 4H_2$
7. $C + 2H_2 = CH_4$
8. $Na_2SO_4 + CaCl_2 = CaSO_4 + 2NaCl$
9. $2C_2H_6 + 7O_2 = 4CO_2 + 6H_2O$
10. $2Al_2O_3 = 4Al + 3O_2$