Question

make the following equations on your desk

	reactants start	products start	reactants final	products final	balanced equation
$ce{h_{2}o_{2} -> h_{2}o + o_{2}}$
$ce{na + o_{2} -> na_{2}o}$
$ce{n_{2} + h_{2} -> nh_{3}}$
$ce{p_{4} + o_{2} -> p_{4}o_{10}}$
$ce{fe + h_{2}o -> fe_{3}o_{4} + h_{2}}$
$ce{c + h_{2} -> ch_{4}}$
$ce{na_{2}so_{4} + cacl_{2} -> caso_{4} + nacl}$
$ce{c_{2}h_{6} + o_{2} -> co_{2} + h_{2}o}$
$ce{al_{2}o_{3} -> al + o_{2}}$

Answer:

To solve these chemical equations, we need to balance them by ensuring the number of each type of atom is the same on both sides of the equation. Here are the balanced equations for each:

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: $2H_2 + O_2
ightarrow 2H_2O$
Now, $H = 4$ (both sides), $O = 2$ (both sides)

Balanced Equation: $2H_2 + O_2 = 2H_2O$

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: $2H_2O_2
ightarrow H_2O + O_2$
Now, $H = 4$, $O = 4$

Step 3: Balance H and O

Multiply $H_2O$ by 2: $2H_2O_2
ightarrow 2H_2O + O_2$
Now, $H = 4$ (both sides), $O = 4$ (both sides)

Balanced Equation: $2H_2O_2 = 2H_2O + O_2$

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, $O = 2$ (both sides), $Na = 4$

Step 3: Balance Na

Multiply $Na$ by 4: $4Na + O_2
ightarrow 2Na_2O$
Now, $Na = 4$ (both sides), $O = 2$ (both sides)

Balanced Equation: $4Na + O_2 = 2Na_2O$

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, $N = 2$ (both sides), $H = 6$

Step 3: Balance H

Multiply $H_2$ by 3: $N_2 + 3H_2
ightarrow 2NH_3$
Now, $N = 2$, $H = 6$ (both sides)

Balanced Equation: $N_2 + 3H_2 = 2NH_3$

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: $P_4 + 5O_2
ightarrow P_4O_{10}$
Now, $O = 10$ (both sides), $P = 4$ (both sides)

Balanced Equation: $P_4 + 5O_2 = P_4O_{10}$

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, $Fe = 3$ (both sides), $O = 1$ (reactants), $O = 4$ (products)

Step 3: Balance O

Multiply $H_2O$ by 4: $3Fe + 4H_2O
ightarrow Fe_3O_4 + H_2$
Now, $O = 4$ (both sides), $H = 8$ (reactants)

Step 4: Balance H

Multiply $H_2$ by 4: $3Fe + 4H_2O = Fe_3O_4 + 4H_2$
Now, $H = 8$ (both sides), $Fe = 3$, $O = 4$ (both sides)

Balanced Equation: $3Fe + 4H_2O = Fe_3O_4 + 4H_2$

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: $C + 2H_2 = CH_4$
Now, $H = 4$ (both sides), $C = 1$ (both sides)

Balanced Equation: $C + 2H_2 = CH_4$

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: $Ca = 1$, $S = 1$, $O = 4$, $Na = 1$, $Cl = 1$

Step 2: Balance Na and Cl

Multiply $NaCl$ by 2: $Na_2SO_4 + CaCl_2 = CaSO_4 + 2NaCl$
Now, $Na = 2$ (both sides), $Cl = 2$ (both sides), others balanced

Balanced Equation: $Na_2SO_4 + CaCl_2 = CaSO_4 + 2NaCl$

9. $\boldsymbol{C_2H_6 + O_2

ightarrow CO_2 + H_2O}$

Step 1: Count atoms

• Reactants: $C = 2$, $H = 6$, $O = 2$
• Products: $C = 1$, $H = 2$, $O = 3$

Step 2: Balance C

Multiply $CO_2$ by 2: $C_2H_6 + O_2
ightarrow 2CO_2 + H_2O$
Now, $C = 2$ (both sides), $H = 6$ (reactants), $H = 2$ (products)

Step 3: Balance H

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

Step 4: Balance O

Multiply $O_2$ by $\frac{7}{2}$: $C_2H_6 + \frac{7}{2}O_2
ightarrow 2CO_2 + 3H_2O$
To eliminate fraction, multiply all by 2: $2C_2H_6 + 7O_2 = 4CO_2 + 6H_2O$
Now, $C = 4$, $H = 12$, $O = 14$ (both sides)

Balanced Equation: $2C_2H_6 + 7O_2 = 4CO_2 + 6H_2O$

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, $Al = 2$ (both sides), $O = 3$ (reactants), $O = 2$ (products)

Step 3: Balance O

Multiply $Al_2O_3$ by 2 and $O_2$ by 3: $2Al_2O_3
ightarrow 4Al + 3O_2$
Now, $Al = 4$ (both sides), $O = 6$ (both sides)

Balanced Equation: $2Al_2O_3 = 4Al + 3O_2$

Summary of 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. $2Al_2O_3 = 4Al + 3O_2$