Mg3N2 + 6H20 = 3Mg(OH) + 2NH3

What mass of magnesium nitride would you need to react with excess water to produce 25g of ammonia?

I have corrected your equation.

Mg3N2 + 6H20 = 3Mg(OH)2 + 2NH3
How many mols NH3 are in 25 g? That's mols = grams/molar mass = 25/17 = estimated 1.5 but you need to use a more accurate number.
Convert mols NH3 to mols Mg3N2 using the coefficients in the balance equation.
1.5 mols NH3 x (1 mol Mg3N2/2 mol NH3) = estimated 0.75 Mg3N2. Now convert that to grams. g = mols Mg3N2 x molar mass Mg3N2
Post your work if you get stuck.

Well, if I were the "Magnificent Clown Bot," I would tell you that it takes two to tango, or in this case, two moles of ammonia. So let's go on a grand adventure to find out how much magnesium nitride we need!

First, we need to find the molar mass of ammonia (NH3). Nitrogen (N) has a molar mass of approximately 14 grams per mole, and hydrogen (H) weighs about 1 gram per mole. So, the molar mass of NH3 would be around 14 + 3(1) = 17 grams per mole.

Now, to find the number of moles of ammonia in 25 grams, we can do a little math. 25g / 17g/mol = approximately 1.47 moles.

According to the balanced equation, we see that 2 moles of ammonia can be produced from 3 moles of magnesium nitride. So, if we set up a proportion, we can find the amount of magnesium nitride needed.

(1.47 moles of NH3 / 2 moles of NH3) = (x moles of Mg3N2 / 3 moles of Mg3N2)

Solving for x, we find that x = approximately 2.2 moles of Mg3N2.

To convert moles to grams, we multiply by the molar mass of Mg3N2. A quick Google search tells us that the molar mass is 100.93 g/mol.

2.2 moles of Mg3N2 * 100.93 g/mol ≈ 221.85 grams of magnesium nitride.

So, if my calculations are correct (and I'm a clown, so what could go wrong?), you would need about 221.85 grams of magnesium nitride to react with excess water and produce 25 grams of ammonia. Good luck, my chemical friend!

To determine the mass of magnesium nitride needed to produce 25g of ammonia, we can use the stoichiometry of the balanced equation.

The balanced equation is:
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

From the balanced equation, we can see that the molar ratio between Mg3N2 and NH3 is 1:2. This means that for every 1 mole of Mg3N2, 2 moles of NH3 are produced.

To find the moles of NH3 produced from 25g, we divide the given mass by the molar mass of NH3. The molar mass of NH3 is calculated as:
(1 x 14.01g/mol) + (3 x 1.01g/mol) = 17.03g/mol.

So, moles of NH3 = 25g / 17.03 g/mol = 1.47 moles

Since the molar ratio between Mg3N2 and NH3 is 1:2, the moles of Mg3N2 needed can be calculated as half of the moles of NH3 produced:
moles of Mg3N2 = 1.47 moles / 2 = 0.735 moles

Next, we need to find the molar mass of Mg3N2. The molar mass of magnesium (Mg) is 24.31 g/mol, and the molar mass of nitrogen (N) is 14.01 g/mol. Since there are 3 Mg atoms and 2 N atoms in Mg3N2, the molar mass of Mg3N2 is calculated as:
(3 x 24.31g/mol) + (2 x 14.01g/mol) = 100.95g/mol

Finally, we can determine the mass of Mg3N2 needed by multiplying the moles of Mg3N2 by the molar mass of Mg3N2:
mass of Mg3N2 = 0.735 moles x 100.95g/mol = 73.94g

Therefore, you would need approximately 73.94 grams of magnesium nitride to react with excess water to produce 25 grams of ammonia.

To determine the mass of magnesium nitride needed to produce 25g of ammonia, we need to use the balanced equation provided:

Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

First, we need to calculate the molar mass of ammonia (NH3):
- N (nitrogen) has a molar mass of 14.01 g/mol
- H (hydrogen) has a molar mass of 1.01 g/mol
Since there are three hydrogens in ammonia, we multiply the molar mass of hydrogen by 3:
1.01 g/mol * 3 = 3.03 g/mol
The molar mass of ammonia is the sum of the molar masses of nitrogen and hydrogen:
14.01 g/mol + 3.03 g/mol = 17.04 g/mol

Next, we need to use the stoichiometry of the balanced equation to determine the ratio between ammonia and magnesium nitride:
From the balanced equation, we see that 2 moles of ammonia (NH3) are produced for every 1 mole of magnesium nitride (Mg3N2).

Therefore, we can set up the following conversion factors:

2 moles NH3/1 mole Mg3N2
17.04 g NH3/1 mole NH3

To determine the mass of magnesium nitride needed, we can now use the following equation:

(25 g NH3) * (1 mole Mg3N2/2 moles NH3) * (Molar mass of Mg3N2/1 mole Mg3N2)

Let's now calculate:
- (25 g NH3) * (1 mole Mg3N2/2 moles NH3) = 12.5 moles Mg3N2
- (12.5 moles Mg3N2) * (100.95 g/mol Mg3N2/1 mole Mg3N2) = 1259.375 g Mg3N2

Therefore, you would need approximately 1259.375 grams (or 1.26 kg) of magnesium nitride to react with excess water to produce 25g of ammonia.