Balance the chemical equations below.

Question

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Here are the solutions for questions 7 and 8.

7. Balance the equations below.

a) $\text{Al} + \text{Cl}_2 \longrightarrow \text{Al}_2\text{Cl}_6$

Balance Al atoms.
There are 2 Al on the right, so put a coefficient of 2 in front of Al on the left.
$2\text{Al} + \text{Cl}_2 \longrightarrow \text{Al}_2\text{Cl}_6$
Balance Cl atoms.
There are 6 Cl on the right, and 2 Cl on the left. Put a coefficient of 3 in front of $\text{Cl}_2$ on the left.
$2\text{Al} + 3\text{Cl}_2 \longrightarrow \text{Al}_2\text{Cl}_6$
The balanced equation is: $\boxed{2\text{Al} + 3\text{Cl}_2 \longrightarrow \text{Al}_2\text{Cl}_6}$
- b) $\text{C}_2\text{H}_6 + \text{O}_2 \longrightarrow \text{CO}_2 + \text{H}_2\text{O}$
Balance C atoms.
There are 2 C on the left, so put a coefficient of 2 in front of $\text{CO}_2$ on the right.
$\text{C}_2\text{H}_6 + \text{O}_2 \longrightarrow 2\text{CO}_2 + \text{H}_2\text{O}$
Balance H atoms.
There are 6 H on the left, so put a coefficient of 3 in front of $\text{H}_2\text{O}$ on the right.
$\text{C}_2\text{H}_6 + \text{O}_2 \longrightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}$
Balance O atoms.
There are $2 \times 2 + 3 \times 1 = 7$ O atoms on the right. There are 2 O atoms on the left. To get 7 O atoms on the left, use a coefficient of $\frac{7}{2}$ for $\text{O}_2$ .
$\text{C}_2\text{H}_6 + \frac{7}{2}\text{O}_2 \longrightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}$

Multiply the entire equation by 2 to remove the fraction.

$2\text{C}_2\text{H}_6 + 7\text{O}_2 \longrightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}$  
The balanced equation is:
$$
\boxed{2\text{C}_2\text{H}_6 + 7\text{O}_2 \longrightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}}
$$

c) $\text{TiCl}_4 + \text{Na} \longrightarrow \text{Ti} + \text{NaCl}$

Balance Cl atoms.
There are 4 Cl on the left, so put a coefficient of 4 in front of NaCl on the right.
$\text{TiCl}_4 + \text{Na} \longrightarrow \text{Ti} + 4\text{NaCl}$
Balance Na atoms.
There are 4 Na on the right, so put a coefficient of 4 in front of Na on the left.
$\text{TiCl}_4 + 4\text{Na} \longrightarrow \text{Ti} + 4\text{NaCl}$
The balanced equation is: $\boxed{\text{TiCl}_4 + 4\text{Na} \longrightarrow \text{Ti} + 4\text{NaCl}}$
- d) $\text{H}_2\text{O}_2 \longrightarrow \text{H}_2\text{O} + \text{O}_2$
Balance O atoms.
There are 2 O on the left and 3 O on the right. To balance, try to make the oxygen atoms on the right an even number by putting a coefficient of 2 in front of $\text{H}_2\text{O}$ .
$\text{H}_2\text{O}_2 \longrightarrow 2\text{H}_2\text{O} + \text{O}_2$
Now there are $2 \times 1 + 2 = 4$ O atoms on the right and 2 O atoms on the left.
Put a coefficient of 2 in front of $\text{H}_2\text{O}_2$ on the left.
$2\text{H}_2\text{O}_2 \longrightarrow 2\text{H}_2\text{O} + \text{O}_2$
Check H atoms.
There are $2 \times 2 = 4$ H atoms on the left and $2 \times 2 = 4$ H atoms on the right. The equation is balanced.
The balanced equation is: $\boxed{2\text{H}_2\text{O}_2 \longrightarrow 2\text{H}_2\text{O} + \text{O}_2}$

8. Complete the following

Compound Element
1. Water Hydrogen
2. Common Salt Sodium
3. Hydrochloric acid Hydrogen
4. Carbon (IV) oxide Carbon
5. Ammonium chloride Nitrogen
6. Sulphuric (VI) acid Sulfur
7. Nitrogen (IV) oxide Nitrogen

Balance the chemical equations below.

Step-by-step solution

Balance Al atoms.

Balance Cl atoms.

Balance C atoms.

Balance H atoms.

Balance O atoms.

Multiply the entire equation by 2 to remove the fraction.

Balance Cl atoms.

Balance Na atoms.

Balance O atoms.

Now there are $2 \times 1 + 2 = 4$ O atoms on the right and 2 O atoms on the left.

Check H atoms.

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More Chemistry Questions

Compound	Element
1. Water	Hydrogen
2. Common Salt	Sodium
3. Hydrochloric acid	Hydrogen
4. Carbon (IV) oxide	Carbon
5. Ammonium chloride	Nitrogen
6. Sulphuric (VI) acid	Sulfur
7. Nitrogen (IV) oxide	Nitrogen

Balance the chemical equations below.

Step-by-step solution

Balance Al atoms.

Balance Cl atoms.

Balance C atoms.

Balance H atoms.

Balance O atoms.

Multiply the entire equation by 2 to remove the fraction.

Balance Cl atoms.

Balance Na atoms.

Balance O atoms.

Now there are 2×1+2=42 \times 1 + 2 = 42×1+2=4 O atoms on the right and 2 O atoms on the left.

Check H atoms.

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Now there are $2 \times 1 + 2 = 4$ O atoms on the right and 2 O atoms on the left.