Ammonia can be produced in a two-step process. First, nitrogen reacts with red-hot magnesium to form magnesium nitride. Then magnesium nitride reacts with water to form magnesium hydroxide and ammonia.

How many grams of magnesium would we have to start with to prepare 15.0 grams of ammonia?

For the balanced chemical formulas are:

3Mg(s) + N2(g) --> Mg3N2(s)
Mg2N2(s) + 6H2O --> 3Mg(OH)2(aq) + 2NH3(aq)

I'm still struggling with finding the grams of Magnesium needed to start with to prepare 15.0 grams of ammonia. I don't know how to do reverse stoichiometry. Please help. Thanks

To determine the amount of magnesium required to produce 15.0 grams of ammonia, we need to consider the stoichiometry of the reaction.

The balanced equation for the reaction between magnesium and nitrogen to form magnesium nitride is:

3 Mg + N2 → Mg3N2

From the equation, we can see that 3 moles of magnesium react with 1 mole of nitrogen to produce 1 mole of magnesium nitride.

The molecular weight of Mg3N2 is:
(3 x atomic weight of Mg) + (2 x atomic weight of N) = (3 x 24.31 g/mol) + (2 x 14.01 g/mol) = 100.72 g/mol

Therefore, 100.72 grams of Mg3N2 is formed from 3 moles of magnesium.

Now, let's calculate the amount of ammonia produced in the second step.

The balanced equation for the reaction between magnesium nitride and water is:

Mg3N2 + 6 H2O → 3 Mg(OH)2 + 2 NH3

From the equation, we can see that 1 mole of magnesium nitride produces 2 moles of ammonia.

The molecular weight of NH3 is:
(1 x atomic weight of N) + (3 x atomic weight of H) = (1 x 14.01 g/mol) + (3 x 1.01 g/mol) = 17.03 g/mol

Therefore, 2 moles of NH3 have a mass of 34.06 grams.

Now, let's calculate the amount of magnesium required to produce 15.0 grams of ammonia.

(15.0 g NH3) x (1 mol NH3 / 34.06 g NH3) x (3 mol Mg3N2 / 2 mol NH3) x (3 mol Mg / 1 mol Mg3N2) = 9.88 g Mg

Therefore, we would need approximately 9.88 grams of magnesium to prepare 15.0 grams of ammonia.

To determine how many grams of magnesium are needed to produce 15.0 grams of ammonia, we need to use stoichiometry.

Let's first write the balanced chemical equation for the formation of ammonia from magnesium:

3Mg + N2 -> Mg3N2

From the equation, we see that three magnesium atoms are required to react with one nitrogen molecule to produce one molecule of magnesium nitride. However, we are given the amount of ammonia (15.0 grams), so we need to convert that into moles.

Using the molar mass of ammonia (NH3), which is 17.03 g/mol, we can calculate the number of moles of ammonia:

moles of ammonia = mass of ammonia / molar mass of ammonia
moles of ammonia = 15.0 g / 17.03 g/mol ≈ 0.881 mol

Now, using the stoichiometry of the balanced equation, we can determine the number of moles of magnesium needed to produce 0.881 mol of ammonia. From the balanced equation, we see that 3 moles of magnesium are required to produce 1 mole of ammonia.

moles of magnesium = 3 × moles of ammonia
moles of magnesium = 3 × 0.881 mol ≈ 2.643 mol

Finally, to calculate the mass of magnesium, we need to convert moles of magnesium to grams using its molar mass, which is 24.31 g/mol.

mass of magnesium = moles of magnesium × molar mass of magnesium
mass of magnesium = 2.643 mol × 24.31 g/mol ≈ 64.0 g

Therefore, approximately 64.0 grams of magnesium would be required to produce 15.0 grams of ammonia in the given two-step process.