calculate the mass of ammonium dicromate which is capable to produce as much nitrogen sufficient to convert 3.6 g of magnesiuminto magnesium nitrides.also calculate the no.of magnesum nitrate prepared???

To tackle this problem, we need to break it down into a few logical steps. First, we need to determine how much nitrogen is required to convert 3.6 g of magnesium into magnesium nitride. Then, we will find out how much ammonium dichromate is needed to produce that amount of nitrogen. Finally, we will calculate the number of magnesium nitrides produced in the reaction.

Step 1: Understanding the Reaction

Magnesium nitride (Mg3N2) forms when magnesium reacts with nitrogen. The balanced chemical equation for this reaction is:

• 3 Mg + N2 → Mg3N2

This equation tells us that 3 moles of magnesium react with 1 mole of nitrogen gas (N2) to produce 1 mole of magnesium nitride.

Step 2: Calculate Moles of Magnesium

First, we need to find out how many moles of magnesium are in 3.6 g. The molar mass of magnesium (Mg) is approximately 24.31 g/mol. We can use the formula:

Number of moles = mass (g) / molar mass (g/mol)

Substituting in the values:

Number of moles of Mg = 3.6 g / 24.31 g/mol ≈ 0.148 moles

Step 3: Determine Moles of Nitrogen Required

From the balanced equation, we know that 3 moles of magnesium require 1 mole of nitrogen. Therefore, to find the moles of nitrogen needed for 0.148 moles of magnesium, we can set up a ratio:

0.148 moles Mg × (1 mole N2 / 3 moles Mg) ≈ 0.0493 moles N2

Step 4: Calculate Mass of Nitrogen

The molar mass of nitrogen gas (N2) is approximately 28.02 g/mol. To find the mass of nitrogen required, we use:

Mass = moles × molar mass

Substituting in the values:

Mass of N2 = 0.0493 moles × 28.02 g/mol ≈ 1.38 g

Step 5: Ammonium Dichromate and Nitrogen Production

Next, we need to find out how much ammonium dichromate ((NH4)2Cr2O7) is required to produce 1.38 g of nitrogen. The decomposition of ammonium dichromate can be represented as:

• (NH4)2Cr2O7 → Cr2O3 + N2 + 4 H2O

This equation shows that 1 mole of ammonium dichromate produces 1 mole of nitrogen gas. The molar mass of ammonium dichromate is approximately 252.07 g/mol. Therefore, to find the moles of ammonium dichromate needed to produce 0.0493 moles of nitrogen:

0.0493 moles N2 × (1 mole (NH4)2Cr2O7 / 1 mole N2) = 0.0493 moles (NH4)2Cr2O7

Step 6: Calculate Mass of Ammonium Dichromate

Now, we can find the mass of ammonium dichromate required:

Mass = moles × molar mass

Mass of (NH4)2Cr2O7 = 0.0493 moles × 252.07 g/mol ≈ 12.43 g

Step 7: Determine Number of Magnesium Nitrides Produced

Finally, we can calculate the number of magnesium nitrides produced. The molar mass of magnesium nitride (Mg3N2) is approximately 100.95 g/mol. Since we produced 0.0493 moles of nitrogen, we can find the moles of magnesium nitride produced:

0.0493 moles N2 × (1 mole Mg3N2 / 1 mole N2) = 0.0493 moles Mg3N2

To find the mass of magnesium nitride produced:

Mass of Mg3N2 = 0.0493 moles × 100.95 g/mol ≈ 4.98 g

In summary, to convert 3.6 g of magnesium into magnesium nitride, you would need approximately 12.43 g of ammonium dichromate, which would yield about 0.0493 moles of magnesium nitride.