Question

homework: balancing chemical compounds
name:
pd:

1. al and s →

1. na and o →

1. cu(ii) and i →

1. fe(iii) and o →

1. co(ii) and s →

1. zn and n →

1. sn(iv) and f →

Explanation:

Question 9: Al and S

Step 1: Determine oxidation states

Aluminum (Al) typically has an oxidation state of +3. Sulfur (S) in binary compounds with metals has an oxidation state of -2.

Step 2: Find the least common multiple (LCM) of the absolute values of the oxidation states

The absolute values are 3 and 2. The LCM of 3 and 2 is 6.

Step 3: Determine the number of each atom

For Al: $\frac{6}{3} = 2$ atoms.
For S: $\frac{6}{2} = 3$ atoms.

Step 4: Write the formula

Combine the atoms: $\text{Al}_2\text{S}_3$ (Aluminum Sulfide).

Question 10: Na and O

Step 1: Determine oxidation states

Sodium (Na) has an oxidation state of +1. Oxygen (O) in binary metal oxides has an oxidation state of -2.

Step 2: Find the LCM of 1 and 2

LCM of 1 and 2 is 2.

Step 3: Determine the number of each atom

For Na: $\frac{2}{1} = 2$ atoms.
For O: $\frac{2}{2} = 1$ atom.

Step 4: Write the formula

Combine the atoms: $\text{Na}_2\text{O}$ (Sodium Oxide).

Question 11: Cu(II) and I

Step 1: Determine oxidation states

Copper(II) has an oxidation state of +2. Iodine (I) in binary compounds with metals has an oxidation state of -1.

Step 2: Find the LCM of 2 and 1

LCM of 2 and 1 is 2.

Step 3: Determine the number of each atom

For Cu: $\frac{2}{2} = 1$ atom.
For I: $\frac{2}{1} = 2$ atoms.

Step 4: Write the formula

Combine the atoms: $\text{CuI}_2$ (Copper(II) Iodide).

Question 12: Fe(III) and O

Answer:

s:

1. $\boldsymbol{\text{Al}_2\text{S}_3}$
2. $\boldsymbol{\text{Na}_2\text{O}}$
3. $\boldsymbol{\text{CuI}_2}$
4. $\boldsymbol{\text{Fe}_2\text{O}_3}$
5. $\boldsymbol{\text{CoS}}$
6. $\boldsymbol{\text{Zn}_3\text{N}_2}$
7. $\boldsymbol{\text{SnF}_4}$