When 4 gm of Mg burnt with `O_(2)` form oxide and on dilution of metal oxide form metal hydroxide which statement is/are correct for above series of reaction?

Options

A. Gram quvalent of metal, metal oxide and metal hydroxide are equal
B. Weight of metal oxide is 6.66 gm
C. Weight of metal hydroxide is 9.66 gm
D. Normaililty of solutioni in 1 litre volume is `14/42`

Answer

To solve the problem, we need to analyze the series of reactions involving magnesium (Mg) and oxygen (O₂) and determine which statements regarding the reactions are correct. Let's break it down step by step.

### Step 1: Write the Reaction for Magnesium Burning in Oxygen
When magnesium burns in oxygen, it forms magnesium oxide (MgO). The balanced chemical equation for this reaction is:

\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \]

### Step 2: Calculate the Molar Masses
- Molar mass of magnesium (Mg) = 24 g/mol
- Molar mass of oxygen (O₂) = 32 g/mol (16 g/mol for each O atom)
- Molar mass of magnesium oxide (MgO) = 40 g/mol (24 g/mol for Mg + 16 g/mol for O)

### Step 3: Determine the Amount of Magnesium and Oxygen Used
Given that 4 grams of magnesium is used, we can calculate the number of moles of magnesium:

\[ \text{Moles of Mg} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \text{ g}}{24 \text{ g/mol}} = \frac{1}{6} \text{ mol} \]

Using the balanced equation, we can find the amount of oxygen required. From the equation, 2 moles of Mg react with 1 mole of O₂. Therefore, for 1 mole of Mg, we need:

\[ \text{Moles of O}_2 = \frac{1}{2} \times \text{Moles of Mg} = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \text{ mol} \]

Now, we can convert moles of O₂ to grams:

\[ \text{Mass of O}_2 = \text{Moles} \times \text{Molar Mass} = \frac{1}{12} \text{ mol} \times 32 \text{ g/mol} = \frac{32}{12} \text{ g} \approx 2.67 \text{ g} \]

### Step 4: Calculate the Mass of Magnesium Oxide Formed
The total mass of magnesium oxide formed can be calculated by adding the mass of magnesium and the mass of oxygen:

\[ \text{Mass of MgO} = \text{Mass of Mg} + \text{Mass of O}_2 = 4 \text{ g} + 2.67 \text{ g} \approx 6.67 \text{ g} \]

### Step 5: Determine the Mass of Magnesium Hydroxide Formed
When magnesium oxide reacts with water, it forms magnesium hydroxide (Mg(OH)₂). The balanced reaction is:

\[ \text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 \]

To find the mass of magnesium hydroxide formed, we need to use the molar masses:
- Molar mass of magnesium hydroxide (Mg(OH)₂) = 58 g/mol (24 g/mol for Mg + 2 × 16 g/mol for O + 2 × 1 g/mol for H)

Using the mass of magnesium oxide formed (6.67 g), we can find the moles of MgO:

\[ \text{Moles of MgO} = \frac{6.67 \text{ g}}{40 \text{ g/mol}} \approx 0.167 \text{ mol} \]

Since 1 mole of MgO produces 1 mole of Mg(OH)₂, the mass of Mg(OH)₂ formed is:

\[ \text{Mass of Mg(OH)}_2 = \text{Moles of MgO} \times \text{Molar Mass of Mg(OH)}_2 = 0.167 \text{ mol} \times 58 \text{ g/mol} \approx 9.67 \text{ g} \]

### Step 6: Calculate Normality of the Solution
Normality (N) is calculated using the formula:

\[ N = \frac{\text{mass of solute (g)}}{\text{equivalent weight (g/equiv)} \times \text{volume (L)}} \]

For magnesium hydroxide, the equivalent weight is the molar mass divided by the number of hydroxide ions (2):

\[ \text{Equivalent weight of Mg(OH)}_2 = \frac{58 \text{ g/mol}}{2} = 29 \text{ g/equiv} \]

Assuming we have 1 L of solution, the normality is:

\[ N = \frac{4 \text{ g}}{29 \text{ g/equiv} \times 1 \text{ L}} \approx 0.138 \text{ N} \]

### Final Conclusion
Now we can evaluate the statements:
1. The gram equivalent of metal, metal oxide, and metal hydroxide are equal.
2. The weight of metal oxide is 6.67 grams.
3. The weight of metal hydroxide is 9.67 grams.
4. The normality of the solution in 1 liter is approximately 0.138 N, which does not match the claim of 14/42.

Thus, the correct statements are:
- Statement 1: True
- Statement 2: True
- Statement 3: True
- Statement 4: False